How to integrate 1/(ax^2+bx+c)?

Make the denominator a perfect square: ax^2+bx+c = a[(x+b/2a)^2 + (4ac-b^2)/4a^2]. Then use standard formulae: ∫dx/(x^2-a^2) = (1/2a)ln|(x-a)/(x+a)|+C; ∫dx/(x^2+a^2) = (1/a)tan^(-1)(x/a)+C; ∫dx/(a^2-x^2) = (1/2a)ln|(a+x)/(a-x)|+C.

What is Euler's substitution?

Euler's substitutions are used to integrate functions of the form ∫R(x, sqrt(ax^2+bx+c))dx. There are three Euler substitutions depending on the sign of a and the discriminant (b^2-4ac). They convert the irrational integrand into a rational function.

What is a reduction formula in integration?

A reduction formula relates an integral with a higher power to an integral with a lower power. For example, ∫sin^n(x)dx = (-1/n)sin^(n-1)(x)cos(x) + ((n-1)/n)∫sin^(n-2)(x)dx. These are derived using integration by parts and reduce the computation of complex integrals to simpler ones.

What is the constant of integration?

Since the derivative of a constant is zero, when we find the anti-derivative, we add an arbitrary constant C called the constant of integration. So ∫f(x)dx = F(x) + C, where C can be any real number. This is why it is called 'indefinite' integration.

Indefinite Integration JEE Notes 2026 | Methods, Formulas, PYQs with Solutions

1

Definition & Basic Concepts

Integration is the reverse process of differentiation. If the derivative of a function \(F(x)\) with respect to \(x\) is \(f(x)\), i.e., \(\frac{d}{dx}F(x) = f(x)\), then we say that \(F(x)\) is an anti-derivative (or primitive) of \(f(x)\), and we write:

Most Repeated PYQ — Definition

The indefinite integral of \(f(x)\) with respect to \(x\) is:

\int f(x)\, dx = F(x) + C

where:

\(\int\) is the integral sign (an elongated S for "sum")
\(f(x)\) is the integrand (the function being integrated)
\(dx\) indicates the variable of integration
\(F(x)\) is the anti-derivative
\(C\) is the constant of integration (arbitrary constant)

1.1 Why is it called "Indefinite"?

Since the derivative of any constant is zero, the anti-derivative of \(f(x)\) is not unique. If \(F(x)\) is an anti-derivative, then \(F(x) + C\) (for any constant \(C\)) is also an anti-derivative. Because \(C\) can take any real value, the integral is "indefinite" — it represents a family of functions rather than a single function.

1.2 Geometrical Interpretation

If we plot \(y = F(x)\) and \(y = F(x) + C\) for different values of \(C\), we get a family of parallel curves (each shifted vertically by \(C\)). The slope at corresponding points is the same (= \(f(x)\)).

The indefinite integral thus represents the family of all curves whose slope at any point \(x\) is \(f(x)\).

1.3 Comparison: Differentiation vs Integration

Aspect	Differentiation	Integration
Process	Finding rate of change	Finding original function (reverse)
Result	Unique function	Family of functions (with +C)
Notation	\(\frac{d}{dx}f(x) = f'(x)\)	\(\int f(x)\,dx = F(x) + C\)
Product rule	\((uv)' = u'v + uv'\)	\(\int u\,dv = uv - \int v\,du\)
Chain rule	\(\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)\)	Substitution method
Always possible?	Yes (for continuous functions)	Not always in closed form

PYQ — JEE Main 2024 (Jan 30 Shift 1)

Question: If \(\int f(x)\,dx = g(x) + C\), then \(\frac{d}{dx}\left[\int f(x)\,dx\right]\) equals:

(a) \(g(x)\)

(b) \(f(x)\)

(c) \(g'(x) + C\)

(d) \(f'(x)\)

Solution:

Given \(\int f(x)\,dx = g(x) + C\).

By definition of integration as reverse of differentiation:

\frac{d}{dx}\left[\int f(x)\,dx\right] = f(x)

This is the Fundamental Theorem of Calculus (Part 1) in essence — differentiation and integration are inverse operations.

\[f(x)\]

Answer: (b) \(f(x)\)

2

Standard Integration Formulae

These are the basic formulae obtained by reversing the standard derivatives. Memorizing these is essential — almost every integration problem reduces to one or more of these.

2.1 Power and Algebraic Functions

Most Repeated PYQ — Power Rule

\int x^n\, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

\int \frac{1}{x}\, dx = \ln|x| + C

\int 1\, dx = x + C

\int a\, dx = ax + C \quad (a \text{ constant})

2.2 Exponential and Logarithmic

\int e^x\, dx = e^x + C

\int a^x\, dx = \frac{a^x}{\ln a} + C, \quad a > 0, a \neq 1

\int \ln x\, dx = x\ln x - x + C

2.3 Trigonometric Functions

Most Repeated PYQ — Trig Integrals

\int \sin x\, dx = -\cos x + C

\int \cos x\, dx = \sin x + C

\int \sec^2 x\, dx = \tan x + C

\int \csc^2 x\, dx = -\cot x + C

\int \sec x \tan x\, dx = \sec x + C

\int \csc x \cot x\, dx = -\csc x + C

\int \tan x\, dx = \ln|\sec x| + C = -\ln|\cos x| + C

\int \cot x\, dx = \ln|\sin x| + C

\int \sec x\, dx = \ln|\sec x + \tan x| + C

\int \csc x\, dx = \ln|\csc x - \cot x| + C = \ln|\tan(x/2)| + C

2.4 Inverse Trigonometric Forms

Most Repeated PYQ — Inverse Trig Forms

\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C

\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C

\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\sec^{-1}\!\left(\frac{x}{a}\right) + C

2.5 Forms Involving Square Roots

Most Repeated PYQ — Surd Forms

\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\!\left|x + \sqrt{x^2 - a^2}\right| + C

\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\!\left|x + \sqrt{x^2 + a^2}\right| + C

\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\!\left|\frac{x - a}{x + a}\right| + C

\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\!\left|\frac{a + x}{a - x}\right| + C

\int \sqrt{x^2 - a^2}\, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\!\left|x + \sqrt{x^2 - a^2}\right| + C

\int \sqrt{x^2 + a^2}\, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\!\left|x + \sqrt{x^2 + a^2}\right| + C

\int \sqrt{a^2 - x^2}\, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right) + C

PYQ — JEE Main 2023 (Apr 10 Shift 2)

Question: The value of \(\int \frac{dx}{x^2 + 4}\) is:

(a) \(\frac{1}{2}\tan^{-1}(x/2) + C\)

(b) \(\tan^{-1}(x/2) + C\)

(c) \(\frac{1}{4}\tan^{-1}(x/2) + C\)

(d) \(\frac{1}{2}\tan^{-1}(2x) + C\)

Solution:

Using the standard formula: \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C\).

Here \(a^2 = 4\), so \(a = 2\).

\int \frac{dx}{x^2 + 4} = \frac{1}{2}\tan^{-1}\!\left(\frac{x}{2}\right) + C

\frac{1}{2}\tan^{-1}\!\left(\frac{x}{2}\right) + C

Answer: (a) \(\frac{1}{2}\tan^{-1}(x/2) + C\)

3

Properties of Indefinite Integrals

Indefinite integrals satisfy several linearity properties and algebraic rules that are essential for manipulating and simplifying integrals before applying standard formulae.

3.1 Linearity Property

Most Repeated PYQ — Linearity

\int [a\,f(x) + b\,g(x)]\, dx = a\int f(x)\,dx + b\int g(x)\,dx

where \(a, b\) are constants.

Special cases:

\(\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx\)
\(\int k\,f(x)\,dx = k\int f(x)\,dx\) (constant multiple)
\(\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx\)

3.2 Integral of Derivative

\int \frac{d}{dx}[f(x)]\, dx = f(x) + C

Integration "undoes" differentiation (up to a constant).

3.3 Derivative of Integral

\frac{d}{dx}\left[\int f(x)\, dx\right] = f(x)

Differentiation "undoes" integration (the constant disappears because derivative of constant is zero).

3.4 Two Integrals are Equal

Equality Condition

If \(\int f(x)\,dx = \int g(x)\,dx\), then \(f(x)\) and \(g(x)\) need not be equal — they differ by a constant: \(f(x) = g(x) + k\) for some constant \(k\).

3.5 Important Note on Constants

Common Mistake

When integrating a sum like \(\int [f(x) + g(x)]\,dx\), you get only ONE constant \(C\), not two separate constants. The constants from each integral combine into a single arbitrary constant.

\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx = F(x) + C_1 + G(x) + C_2 = F(x) + G(x) + C

PYQ — JEE Main 2022 (June 28 Shift 1)

Question: Evaluate \(\int (3x^2 - 2\cos x + 4e^x + \frac{5}{x})\, dx\).

Solution:

Using linearity property, integrate term by term:

\int 3x^2\,dx = x^3

\int 2\cos x\,dx = 2\sin x

\int 4e^x\,dx = 4e^x

\int \frac{5}{x}\,dx = 5\ln|x|

Combining with single constant:

x^3 - 2\sin x + 4e^x + 5\ln|x| + C

Answer: \(x^3 - 2\sin x + 4e^x + 5\ln|x| + C\)

4

Integration by Substitution

When the integrand contains a function and its derivative (or a function whose derivative is also present up to a constant), we use the method of substitution. This is the reverse of the chain rule of differentiation and is the most commonly used integration technique.

4.1 The Substitution Rule

Most Repeated PYQ — Substitution

If we substitute \(t = g(x)\) where \(g\) is differentiable, then:

\int f(g(x))\cdot g'(x)\, dx = \int f(t)\, dt = F(t) + C = F(g(x)) + C

where \(t = g(x)\) and \(dt = g'(x)\,dx\).

4.2 Steps for Substitution

Identify a part of the integrand as \(t = g(x)\) such that \(g'(x)\,dx\) (or a multiple of it) is also present.
Compute \(dt = g'(x)\,dx\).
Substitute to convert the integral entirely in terms of \(t\).
Integrate with respect to \(t\) using standard formulae.
Replace \(t\) back with \(g(x)\) to get the final answer.

4.3 Standard Substitutions

Most Repeated PYQ — Standard Substitutions

Integrand Form	Substitution
\(\int f(ax+b)\,dx\)	\(t = ax+b\)
\(\int \frac{f'(x)}{f(x)}\,dx\)	\(t = f(x)\) → gives \(\ln\|f(x)\| + C\)
\(\int f(x)^n \cdot f'(x)\,dx\)	\(t = f(x)\) → gives \(\frac{f(x)^{n+1}}{n+1} + C\)
\(\int \frac{dx}{x^2+a^2}\)	\(x = a\tan\theta\)
\(\int \frac{dx}{\sqrt{a^2-x^2}}\)	\(x = a\sin\theta\)
\(\int \frac{dx}{\sqrt{x^2-a^2}}\)	\(x = a\sec\theta\)
\(\int \frac{dx}{\sqrt{x^2+a^2}}\)	\(x = a\tan\theta\)
\(\int \frac{dx}{x\sqrt{x^2-a^2}}\)	\(x = a\sec\theta\)
\(\sqrt{a^2-x^2}\) in integrand	\(x = a\sin\theta\)
\(\sqrt{x^2-a^2}\) in integrand	\(x = a\sec\theta\)
\(\sqrt{x^2+a^2}\) in integrand	\(x = a\tan\theta\)

4.4 Important Special Cases

Must-Remember Results

\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C

\int [f(x)]^n \cdot f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + C, \quad n \neq -1

\int e^{f(x)} \cdot f'(x)\,dx = e^{f(x)} + C

\int \frac{f'(x)}{\sqrt{f(x)}}\,dx = 2\sqrt{f(x)} + C

PYQ — JEE Main 2024 (Apr 5 Shift 2)

Question: Evaluate \(\int \frac{x}{x^2 + 1}\, dx\).

(a) \(\ln(x^2+1) + C\)

(b) \(\frac{1}{2}\ln(x^2+1) + C\)

(c) \(\tan^{-1}(x) + C\)

(d) \(\ln|x^2+1| + C\)

Solution:

Let \(t = x^2 + 1\), then \(dt = 2x\,dx\), so \(x\,dx = \frac{dt}{2}\).

\int \frac{x\,dx}{x^2+1} = \int \frac{dt/2}{t} = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\ln|t| + C

Substituting back \(t = x^2 + 1\) (always positive, so |·| not needed):

\frac{1}{2}\ln(x^2+1) + C

Answer: (b) \(\frac{1}{2}\ln(x^2+1) + C\)

5

Integration of Trigonometric Functions

Many integrals involving trigonometric functions can be evaluated by first using trigonometric identities to simplify the integrand, then applying standard formulae.

5.1 Powers of Sine and Cosine

Most Repeated PYQ — Powers of sin/cos

Odd power of sine: \(\int \sin^{2n+1} x\, dx\). Write \(\sin^{2n+1}x = (1-\cos^2 x)^n \sin x\) and substitute \(t = \cos x\).

Odd power of cosine: \(\int \cos^{2n+1} x\, dx\). Write \(\cos^{2n+1}x = (1-\sin^2 x)^n \cos x\) and substitute \(t = \sin x\).

Even powers: Use half-angle formulae \(\sin^2 x = \frac{1-\cos 2x}{2}\), \(\cos^2 x = \frac{1+\cos 2x}{2}\).

5.2 Products of Sines and Cosines

Most Repeated PYQ — Product-to-Sum

\int \sin mx \sin nx\, dx = \frac{\sin(m-n)x}{2(m-n)} - \frac{\sin(m+n)x}{2(m+n)} + C

\int \cos mx \cos nx\, dx = \frac{\sin(m-n)x}{2(m-n)} + \frac{\sin(m+n)x}{2(m+n)} + C

\int \sin mx \cos nx\, dx = -\frac{\cos(m+n)x}{2(m+n)} - \frac{\cos(m-n)x}{2(m-n)} + C

5.3 Important Trigonometric Identities for Integration

\(\sin^2 x = \frac{1 - \cos 2x}{2}\)
\(\cos^2 x = \frac{1 + \cos 2x}{2}\)
\(\sin^3 x = \frac{3\sin x - \sin 3x}{4}\)
\(\cos^3 x = \frac{3\cos x + \cos 3x}{4}\)
\(2\sin A \cos B = \sin(A+B) + \sin(A-B)\)
\(2\cos A \cos B = \cos(A+B) + \cos(A-B)\)
\(2\sin A \sin B = \cos(A-B) - \cos(A+B)\)
\(\sin^4 x = \frac{3 - 4\cos 2x + \cos 4x}{8}\)
\(\cos^4 x = \frac{3 + 4\cos 2x + \cos 4x}{8}\)

5.4 Forms Involving tan and cot

Useful Results

\int \tan^n x\, dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2} x\, dx

This uses \(\tan^2 x = \sec^2 x - 1\) and is a reduction formula.

PYQ — JEE Main 2024 (Jan 31 Shift 2)

Question: Evaluate \(\int \sin^3 x \cos^2 x\, dx\).

Solution:

Since power of sine is odd (3), write \(\sin^3 x = \sin^2 x \cdot \sin x = (1-\cos^2 x)\sin x\).

\int \sin^3 x \cos^2 x\, dx = \int (1-\cos^2 x)\cos^2 x \cdot \sin x\, dx

Substitute \(t = \cos x\), \(dt = -\sin x\, dx\):

= -\int (1-t^2)t^2\, dt = -\int (t^2 - t^4)\, dt

= -\left[\frac{t^3}{3} - \frac{t^5}{5}\right] + C = \frac{t^5}{5} - \frac{t^3}{3} + C

Substituting back \(t = \cos x\):

\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C

Answer: \(\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C\)

6

Some Special Integrals

These are non-standard but frequently occurring integrals that should be memorized. They are derived using substitution and/or integration by parts.

6.1 Exponential-Linear Forms

Most Repeated PYQ — Exponential Forms

\int e^{ax}\, dx = \frac{e^{ax}}{a} + C

\int e^{ax} \sin bx\, dx = \frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C

\int e^{ax} \cos bx\, dx = \frac{e^{ax}}{a^2+b^2}(a\cos bx + b\sin bx) + C

6.2 Special Form: f(x) + f'(x) Pattern

Most Repeated PYQ — f + f' Pattern

\int e^x [f(x) + f'(x)]\, dx = e^x f(x) + C

This is one of the most important and frequently asked results in JEE. The idea: if the integrand is \(e^x\) times (a function + its derivative), the answer is simply \(e^x\) times the function.

General form: \(\int e^{g(x)}[f(x)g'(x) + f'(x)]\, dx = e^{g(x)}f(x) + C\)

6.3 Logarithmic Forms

\int \ln x\, dx = x\ln x - x + C

\int (\ln x)^2\, dx = x(\ln x)^2 - 2x\ln x + 2x + C

\int \frac{\ln x}{x}\, dx = \frac{(\ln x)^2}{2} + C

\int \frac{dx}{x\ln x} = \ln|\ln x| + C

6.4 Inverse Trigonometric Functions

\int \sin^{-1} x\, dx = x\sin^{-1} x + \sqrt{1-x^2} + C

\int \cos^{-1} x\, dx = x\cos^{-1} x - \sqrt{1-x^2} + C

\int \tan^{-1} x\, dx = x\tan^{-1} x - \frac{1}{2}\ln(1+x^2) + C

\int \cot^{-1} x\, dx = x\cot^{-1} x + \frac{1}{2}\ln(1+x^2) + C

PYQ — JEE Main 2023 (Jan 31 Shift 2)

Question: Evaluate \(\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx\).

Solution:

Write the integrand in the form \(e^x[f(x) + f'(x)]\).

Let \(f(x) = \frac{1}{x}\). Then \(f'(x) = -\frac{1}{x^2}\).

So \(f(x) + f'(x) = \frac{1}{x} - \frac{1}{x^2}\) ✓

Using the formula \(\int e^x[f(x) + f'(x)]\, dx = e^x f(x) + C\):

e^x \cdot \frac{1}{x} + C = \frac{e^x}{x} + C

Answer: \(\frac{e^x}{x} + C\)

7

Integration by Parts (ILATE Rule)

When the integrand is a product of two different types of functions, we use integration by parts. This is the reverse of the product rule of differentiation and is one of the most frequently tested methods in JEE.

7.1 The Formula

Most Repeated PYQ — Integration by Parts

\int u\, dv = uv - \int v\, du

where \(u\) and \(v\) are functions of \(x\), and we choose which function to call \(u\) (to differentiate) and which to call \(dv\) (to integrate).

7.2 The ILATE Rule

Most Repeated PYQ — ILATE Rule

To decide which function to take as \(u\) (the one to differentiate), follow the priority order ILATE:

I — Inverse trigonometric functions (\(\sin^{-1}x, \cos^{-1}x, \tan^{-1}x\), etc.)
L — Logarithmic functions (\(\ln x, \log x\), etc.)
A — Algebraic functions (polynomials: \(x, x^2, x^n\), etc.)
T — Trigonometric functions (\(\sin x, \cos x, \tan x\), etc.)
E — Exponential functions (\(e^x, a^x\), etc.)

The function appearing earlier in ILATE is chosen as \(u\) (to differentiate), and the other as \(dv\) (to integrate).

7.3 Special Forms Using Integration by Parts

\int e^x[f(x) + f'(x)]\, dx = e^x f(x) + C

\int x^n e^{ax}\, dx = \frac{x^n e^{ax}}{a} - \frac{n}{a}\int x^{n-1}e^{ax}\, dx

\int x^n \sin ax\, dx = -\frac{x^n \cos ax}{a} + \frac{n}{a}\int x^{n-1}\cos ax\, dx

\int x^n \cos ax\, dx = \frac{x^n \sin ax}{a} - \frac{n}{a}\int x^{n-1}\sin ax\, dx

\int x^n \ln x\, dx = \frac{x^{n+1}}{n+1}\left[\ln x - \frac{1}{n+1}\right] + C

7.4 Cyclic Integrals

Cyclic Pattern

Some integrals, when integrated by parts twice, return to the original integral. We then solve algebraically:

\int e^{ax}\sin bx\, dx = \frac{e^{ax}}{a^2+b^2}(a\sin bx - b\cos bx) + C

\int e^{ax}\cos bx\, dx = \frac{e^{ax}}{a^2+b^2}(a\cos bx + b\sin bx) + C

PYQ — JEE Main 2024 (Apr 6 Shift 1)

Question: Evaluate \(\int x \cos x\, dx\).

(a) \(x\sin x + \cos x + C\)

(b) \(x\sin x - \cos x + C\)

(c) \(-x\sin x + \cos x + C\)

(d) \(-x\sin x - \cos x + C\)

Solution:

By ILATE: \(u = x\) (Algebraic), \(dv = \cos x\, dx\) (Trigonometric).

Then \(du = dx\), \(v = \sin x\).

Using \(\int u\,dv = uv - \int v\,du\):

\int x\cos x\, dx = x\sin x - \int \sin x\, dx

= x\sin x - (-\cos x) + C = x\sin x + \cos x + C

x\sin x + \cos x + C

Answer: (a) \(x\sin x + \cos x + C\)

8

Integration by Partial Fractions

When the integrand is a rational function \(P(x)/Q(x)\) where degree of \(P\) is less than degree of \(Q\), we decompose it into simpler fractions called partial fractions, then integrate each term.

8.1 Proper vs Improper Rational Function

Proper: degree of numerator < degree of denominator.

Improper: degree of numerator ≥ degree of denominator. For improper, first do polynomial long division to write it as: quotient + remainder/denominator, then apply partial fractions to the remainder part.

8.2 The Four Cases of Partial Fractions

Most Repeated PYQ — Partial Fractions Cases

Case 1: Distinct Linear Factors

If \(Q(x) = (x-a)(x-b)(x-c)\) with distinct \(a, b, c\):

\frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}

Case 2: Repeated Linear Factors

If \(Q(x) = (x-a)^n\):

\frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}

Case 3: Irreducible Quadratic Factor (non-repeated)

If \(Q(x)\) contains \((ax^2+bx+c)\) where \(b^2-4ac < 0\):

\frac{P(x)}{(px+q)(ax^2+bx+c)} = \frac{A}{px+q} + \frac{Bx+C}{ax^2+bx+c}

Case 4: Repeated Irreducible Quadratic

\frac{P(x)}{(ax^2+bx+c)^n} = \frac{A_1 x + B_1}{ax^2+bx+c} + \frac{A_2 x + B_2}{(ax^2+bx+c)^2} + \cdots

8.3 Heaviside Cover-Up Method (Shortcut)

Shortcut for Case 1

For \(\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}\):

A = \frac{P(a)}{(a-b)(a-c)}, \quad B = \frac{P(b)}{(b-a)(b-c)}, \quad C = \frac{P(c)}{(c-a)(c-b)}

This is obtained by multiplying both sides by \((x-a)\) and substituting \(x = a\).

8.4 Integration Results from Partial Fractions

\int \frac{A}{x-a}\, dx = A\ln|x-a| + C

\int \frac{A}{(x-a)^n}\, dx = \frac{-A}{(n-1)(x-a)^{n-1}} + C, \quad n \neq 1

\int \frac{Bx+C}{ax^2+bx+c}\, dx = \text{(split into log + arctan forms)}

PYQ — JEE Main 2022 (July 25 Shift 1)

Question: Evaluate \(\int \frac{dx}{x^2 - 5x + 6}\).

Solution:

Factor: \(x^2 - 5x + 6 = (x-2)(x-3)\).

Partial fractions: \(\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}\).

Using cover-up method: \(A = \frac{1}{2-3} = -1\), \(B = \frac{1}{3-2} = 1\).

\int \frac{dx}{(x-2)(x-3)} = \int\left(\frac{-1}{x-2} + \frac{1}{x-3}\right)dx

= -\ln|x-2| + \ln|x-3| + C = \ln\left|\frac{x-3}{x-2}\right| + C

\ln\left|\frac{x-3}{x-2}\right| + C

Answer: \(\ln\left|\frac{x-3}{x-2}\right| + C\)

9

Integration of 1/(ax²+bx+c) Form

This is one of the most frequently tested forms in JEE. The strategy: complete the square in the denominator, then use a standard formula based on the sign of the discriminant.

9.1 Method: Complete the Square

Most Repeated PYQ — 1/(ax²+bx+c)

Write \(ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}\right]\).

Let \(D = b^2 - 4ac\) (discriminant). The form of the answer depends on the sign of \(D\):

If \(D < 0\) (i.e., \(4ac - b^2 > 0\)): \(\tan^{-1}\) form
If \(D > 0\): \(\ln\) form (partial fractions also possible)
If \(D = 0\): power form

9.2 The Three Results

Most Repeated PYQ — Three Forms

Case 1: \(D < 0\) (denominator always positive)

\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\!\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right) + C

Case 2: \(D > 0\) (denominator has real roots)

\int \frac{dx}{ax^2+bx+c} = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C

Case 3: \(D = 0\) (perfect square)

\int \frac{dx}{ax^2+bx+c} = \frac{-1}{a\left(x + \frac{b}{2a}\right)} + C = \frac{-2}{2ax+b} + C

9.3 Summary of Standard Results

\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C

\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C

\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C

PYQ — JEE Main 2021 (Aug 26 Shift 1)

Question: Evaluate \(\int \frac{dx}{x^2 + 2x + 5}\).

Solution:

Complete the square: \(x^2 + 2x + 5 = (x+1)^2 + 4 = (x+1)^2 + 2^2\).

So \(\int \frac{dx}{x^2+2x+5} = \int \frac{dx}{(x+1)^2 + 2^2}\).

Using \(\int \frac{dt}{t^2+a^2} = \frac{1}{a}\tan^{-1}(t/a) + C\) with \(t = x+1, a = 2\):

= \frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C

\frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C

Answer: \(\frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C\)

10

Integration of (px+q)/(ax²+bx+c) Form

When the numerator is a linear function and denominator is a quadratic, we split the numerator into two parts: one proportional to the derivative of the denominator (giving \(\ln\) form) and the other a constant (giving \(\tan^{-1}\) or \(\ln\) form).

10.1 The Method

Most Repeated PYQ — (px+q)/(ax²+bx+c)

Write numerator \(px + q\) as:

px + q = \lambda(2ax + b) + \mu

where \(\lambda = \frac{p}{2a}\) and \(\mu = q - \frac{pb}{2a}\).

Then:

\int \frac{px+q}{ax^2+bx+c}\,dx = \lambda\int \frac{2ax+b}{ax^2+bx+c}\,dx + \mu\int \frac{dx}{ax^2+bx+c}

First integral: \(\lambda\ln|ax^2+bx+c|\) (since numerator = derivative of denominator).

Second integral: use the method from Chapter 9.

10.2 Step-by-Step Procedure

Find derivative of denominator: \(\frac{d}{dx}(ax^2+bx+c) = 2ax+b\).
Express numerator: \(px + q = A(2ax+b) + B\).
Solve: \(A = \frac{p}{2a}\), \(B = q - \frac{pb}{2a}\).
Split integral: \(\int \frac{A(2ax+b)}{ax^2+bx+c}dx + \int \frac{B}{ax^2+bx+c}dx\).
First part → \(A\ln|ax^2+bx+c| + C_1\).
Second part → complete the square and use standard form.

PYQ — JEE Main 2024 (Apr 8 Shift 1)

Question: Evaluate \(\int \frac{2x+3}{x^2+2x+5}\, dx\).

Solution:

Derivative of denominator: \(\frac{d}{dx}(x^2+2x+5) = 2x+2\).

Express \(2x+3 = A(2x+2) + B\):

Comparing: \(A = 1\), then \(2A + B = 3 \Rightarrow B = 1\).

So \(2x+3 = (2x+2) + 1\).

\int \frac{2x+3}{x^2+2x+5}dx = \int \frac{2x+2}{x^2+2x+5}dx + \int \frac{dx}{x^2+2x+5}

First integral: \(\ln|x^2+2x+5| + C_1\) (numerator = derivative of denominator).

Second integral: \(x^2+2x+5 = (x+1)^2 + 4\), so:

\int \frac{dx}{(x+1)^2+4} = \frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C_2

Combining:

\ln(x^2+2x+5) + \frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C

Answer: \(\ln(x^2+2x+5) + \frac{1}{2}\tan^{-1}\!\left(\frac{x+1}{2}\right) + C\)

11

Integration of 1/√(ax²+bx+c) Form

When the denominator is a square root of a quadratic, we again complete the square, then use a standard formula based on the signs of \(a\) and the completed square.

11.1 Method and Results

Most Repeated PYQ — 1/√(ax²+bx+c)

Complete the square: \(ax^2+bx+c = a\left[\left(x+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a^2}\right]\). Three cases based on sign of \(a\) and discriminant:

Case 1: \(a > 0\) (after completing square, form is \(u^2 + k^2\))

\int \frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\ln\left|2ax+b+2\sqrt{a}\sqrt{ax^2+bx+c}\right| + C

Or equivalently: \(\frac{1}{\sqrt{a}}\ln\left|x+\frac{b}{2a}+\sqrt{x^2+\frac{b}{a}x+\frac{c}{a}}\right| + C\)

Case 2: \(a < 0\) (form is \(k^2 - u^2\))

\int \frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{-a}}\sin^{-1}\!\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right) + C

11.2 Standard Results to Memorize

\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C

\int \frac{dx}{\sqrt{x^2+a^2}} = \ln\left|x+\sqrt{x^2+a^2}\right| + C

\int \frac{dx}{\sqrt{x^2-a^2}} = \ln\left|x+\sqrt{x^2-a^2}\right| + C

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Question: Evaluate \(\int \frac{dx}{\sqrt{4-x^2}}\).

(a) \(\sin^{-1}(x/2) + C\)

(b) \(\frac{1}{2}\sin^{-1}(x/2) + C\)

(c) \(\ln|x+\sqrt{4-x^2}| + C\)

(d) \(2\sin^{-1}(x/2) + C\)

Solution:

Using \(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C\) with \(a^2 = 4, a = 2\):

\int \frac{dx}{\sqrt{4-x^2}} = \sin^{-1}\!\left(\frac{x}{2}\right) + C

\sin^{-1}\!\left(\frac{x}{2}\right) + C

Answer: (a) \(\sin^{-1}(x/2) + C\)

12

Integration of (px+q)/√(ax²+bx+c) Form

Similar to the non-surd case, we split the linear numerator into a part proportional to the derivative of the quadratic (giving surd form) and a constant (giving \(\sin^{-1}\) or \(\ln\) form).

12.1 The Method

Most Repeated PYQ — (px+q)/√(ax²+bx+c)

Write \(px + q = A(2ax+b) + B\), where \(A = \frac{p}{2a}\), \(B = q - \frac{pb}{2a}\).

\int \frac{px+q}{\sqrt{ax^2+bx+c}}\,dx = A\int \frac{2ax+b}{\sqrt{ax^2+bx+c}}\,dx + B\int \frac{dx}{\sqrt{ax^2+bx+c}}

First integral: substitute \(t = ax^2+bx+c\), \(dt = (2ax+b)dx\):

A\int \frac{dt}{\sqrt{t}} = 2A\sqrt{t} = 2A\sqrt{ax^2+bx+c}

Second integral: use method from Chapter 11.

PYQ — JEE Main 2021 (Jul 27 Shift 2)

Question: Evaluate \(\int \frac{x+1}{\sqrt{x^2+4}}\, dx\).

Solution:

Split: \(x+1 = \frac{1}{2}(2x) + 1\).

\int \frac{x+1}{\sqrt{x^2+4}}dx = \frac{1}{2}\int \frac{2x\,dx}{\sqrt{x^2+4}} + \int \frac{dx}{\sqrt{x^2+4}}

First: Let \(t = x^2+4\), \(dt = 2x\,dx\): \(\frac{1}{2}\int \frac{dt}{\sqrt{t}} = \sqrt{t} = \sqrt{x^2+4}\).

Second: \(\int \frac{dx}{\sqrt{x^2+4}} = \ln|x+\sqrt{x^2+4}|\).

\sqrt{x^2+4} + \ln\left|x+\sqrt{x^2+4}\right| + C

Answer: \(\sqrt{x^2+4} + \ln\left|x+\sqrt{x^2+4}\right| + C\)

13

Integration of 1/(Linear × Quadratic)

For integrals of the form \(\int \frac{dx}{(ax+b)(px^2+qx+r)}\), we use partial fractions with one linear and one irreducible quadratic factor.

13.1 The Decomposition

Most Repeated PYQ — Linear × Quadratic

\frac{1}{(ax+b)(px^2+qx+r)} = \frac{A}{ax+b} + \frac{Bx+C}{px^2+qx+r}

Multiply both sides by the denominator and compare coefficients to find \(A, B, C\).

The first part integrates to a logarithm, the second part is handled by the method of Chapter 10.

13.2 General Strategy

Write the partial fraction decomposition.
Solve for constants \(A, B, C\) by comparing coefficients or substituting values.
\(\int \frac{A}{ax+b}\,dx = \frac{A}{a}\ln|ax+b|\).
\(\int \frac{Bx+C}{px^2+qx+r}\,dx\): split using derivative method (Chapter 10).

PYQ — JEE Main 2020 (Sep 5 Shift 1)

Question: Evaluate \(\int \frac{dx}{(x+1)(x^2+1)}\).

Solution:

Partial fractions: \(\frac{1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}\).

Multiply: \(1 = A(x^2+1) + (Bx+C)(x+1)\).

Set \(x = -1\): \(1 = A(2) \Rightarrow A = \frac{1}{2}\).

Compare \(x^2\): \(0 = A + B \Rightarrow B = -\frac{1}{2}\).

Compare constant: \(1 = A + C \Rightarrow C = \frac{1}{2}\).

\int \frac{dx}{(x+1)(x^2+1)} = \frac{1}{2}\int \frac{dx}{x+1} + \int \frac{-x/2+1/2}{x^2+1}dx

= \frac{1}{2}\ln|x+1| - \frac{1}{2}\int\frac{x\,dx}{x^2+1} + \frac{1}{2}\int\frac{dx}{x^2+1}

= \frac{1}{2}\ln|x+1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x + C

\frac{1}{2}\ln|x+1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x + C

Answer: \(\frac{1}{2}\ln|x+1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x + C\)

14

Trigonometric Substitutions

When the integrand contains expressions like \(\sqrt{a^2-x^2}\), \(\sqrt{x^2+a^2}\), or \(\sqrt{x^2-a^2}\), we use trigonometric substitutions to eliminate the square root.

14.1 The Three Standard Substitutions

Most Repeated PYQ — Trig Substitutions

Expression	Substitution	Identity Used
\(\sqrt{a^2-x^2}\)	\(x = a\sin\theta\)	\(1-\sin^2\theta = \cos^2\theta\)
\(\sqrt{a^2+x^2}\)	\(x = a\tan\theta\)	\(1+\tan^2\theta = \sec^2\theta\)
\(\sqrt{x^2-a^2}\)	\(x = a\sec\theta\)	\(\sec^2\theta-1 = \tan^2\theta\)

14.2 Detailed Results of Each Substitution

If \(x = a\sin\theta\): \(dx = a\cos\theta\,d\theta\), \(\sqrt{a^2-x^2} = a\cos\theta\).

If \(x = a\tan\theta\): \(dx = a\sec^2\theta\,d\theta\), \(\sqrt{a^2+x^2} = a\sec\theta\).

If \(x = a\sec\theta\): \(dx = a\sec\theta\tan\theta\,d\theta\), \(\sqrt{x^2-a^2} = a\tan\theta\).

14.3 Other Useful Substitutions

Additional Substitutions

For \(a^2 - x^2\) in denominator (not surd): sometimes \(x = a\sin\theta\) or partial fractions.
For \(\frac{1}{x\sqrt{x^2-a^2}}\): use \(x = a\sec\theta\), result is \(\frac{1}{a}\sec^{-1}(x/a)\).
\(x = a\cosh t\) for \(\sqrt{x^2-a^2}\) (hyperbolic, less common in JEE).
\(x = a\sinh t\) for \(\sqrt{x^2+a^2}\) (hyperbolic).

PYQ — JEE Main 2019 (Apr 10 Shift 1)

Question: Evaluate \(\int \frac{dx}{\sqrt{9-4x^2}}\).

Solution:

Rewrite: \(\sqrt{9-4x^2} = \sqrt{4\left(\frac{9}{4}-x^2\right)} = 2\sqrt{\frac{9}{4}-x^2}\).

\int \frac{dx}{\sqrt{9-4x^2}} = \frac{1}{2}\int \frac{dx}{\sqrt{(3/2)^2 - x^2}}

Using \(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C\) with \(a = 3/2\):

= \frac{1}{2}\sin^{-1}\!\left(\frac{x}{3/2}\right) + C = \frac{1}{2}\sin^{-1}\!\left(\frac{2x}{3}\right) + C

\frac{1}{2}\sin^{-1}\!\left(\frac{2x}{3}\right) + C

Answer: \(\frac{1}{2}\sin^{-1}\!\left(\frac{2x}{3}\right) + C\)

15

Euler's Substitutions

Euler's substitutions are used to integrate functions of the form \(\int R\left(x, \sqrt{ax^2+bx+c}\right)dx\) where \(R\) is a rational function. There are three Euler substitutions depending on the nature of the quadratic.

15.1 First Euler Substitution (when \(a > 0\))

Important — Euler's First

\sqrt{ax^2+bx+c} = t \pm x\sqrt{a}

Squaring and solving for \(x\) gives \(x\) as a rational function of \(t\), converting the integral into a rational function.

15.2 Second Euler Substitution (when \(c > 0\))

Important — Euler's Second

\sqrt{ax^2+bx+c} = xt \pm \sqrt{c}

Used when \(c > 0\), i.e., the quadratic is positive at \(x = 0\).

15.3 Third Euler Substitution (when discriminant \(b^2-4ac > 0\))

Important — Euler's Third

If the quadratic \(ax^2+bx+c\) has real roots \(\alpha\) and \(\beta\) (i.e., \(b^2-4ac > 0\)), then:

\sqrt{ax^2+bx+c} = (x-\alpha)t \quad \text{or} \quad \sqrt{a(x-\alpha)(x-\beta)} = (x-\alpha)t

This gives \(x = \frac{\alpha t^2 - a\beta}{t^2 - a}\), a rational function of \(t\).

15.4 When to Use Which?

\(a > 0\): Use first substitution.
\(c > 0\): Use second substitution.
\(b^2 - 4ac > 0\): Use third substitution.
Often more than one is applicable; choose the one that simplifies the problem most.
In JEE, Euler substitutions are rarely directly required — standard trig substitutions usually suffice.

PYQ — JEE Advanced 2018 (Apr 9 Shift 2)

Question: Evaluate \(\int \frac{dx}{x+\sqrt{x^2-1}}\) using an appropriate substitution.

Solution:

Use Euler's first substitution (\(a = 1 > 0\)): \(\sqrt{x^2-1} = t - x\), so \(t = x + \sqrt{x^2-1}\).

Then the integral becomes \(\int \frac{dx}{t}\).

From \(t = x + \sqrt{x^2-1}\): \(\sqrt{x^2-1} = t - x\), squaring: \(x^2 - 1 = t^2 - 2tx + x^2\).

So \(2tx = t^2 + 1\), giving \(x = \frac{t^2+1}{2t}\) and \(dx = \frac{t^2-1}{2t^2}dt\).

\int \frac{dx}{t} = \int \frac{1}{t}\cdot\frac{t^2-1}{2t^2}dt = \int \frac{t^2-1}{2t^3}dt = \frac{1}{2}\int\left(\frac{1}{t}-\frac{1}{t^3}\right)dt

= \frac{1}{2}\left[\ln|t| + \frac{1}{2t^2}\right] + C

Substituting back \(t = x + \sqrt{x^2-1}\):

\frac{1}{2}\ln\left|x+\sqrt{x^2-1}\right| + \frac{1}{4\left(x+\sqrt{x^2-1}\right)^2} + C

Answer: \(\frac{1}{2}\ln\left|x+\sqrt{x^2-1}\right| + \frac{1}{4\left(x+\sqrt{x^2-1}\right)^2} + C\)

16

Weierstrass Substitution (Universal Trig Substitution)

The Weierstrass substitution \(t = \tan(x/2)\) converts any rational function of \(\sin x\) and \(\cos x\) into a rational function of \(t\), which can then be integrated using partial fractions.

16.1 The Substitution

Most Repeated PYQ — Weierstrass

Let \(t = \tan(x/2)\). Then:

\sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2\,dt}{1+t^2}

Also: \(\tan x = \frac{2t}{1-t^2}\), \(\sec x = \frac{1+t^2}{1-t^2}\), \(\csc x = \frac{1+t^2}{2t}\).

16.2 When to Use Weierstrass

Usage Guidelines

Use when integrand is a rational function of \(\sin x\) and \(\cos x\) that doesn't yield to simpler methods.
Particularly useful when the integrand has both \(\sin x\) and \(\cos x\) with no obvious substitution.
Caution: It often leads to lengthy algebra. Try simpler methods first (identities, substitution).
Special cases: if integrand is invariant under \(x \to -x\) (even in cos), try \(t = \sin x\); if invariant under \(x \to \pi - x\), try \(t = \cos x\); if invariant under \(x \to \pi + x\), try \(t = \tan x\).

16.3 Derivation of the Substitution

Using double angle formulas with \(t = \tan(x/2)\):

\(\sin x = 2\sin(x/2)\cos(x/2) = 2\tan(x/2)\cos^2(x/2) = \frac{2t}{1+t^2}\)
\(\cos x = \cos^2(x/2) - \sin^2(x/2) = \frac{1-t^2}{1+t^2}\)
\(dx = \frac{2\,dt}{1+t^2}\) (from \(x = 2\tan^{-1}t\))

PYQ — JEE Main 2020 (Oct 5 Shift 1)

Question: Evaluate \(\int \frac{dx}{1+\sin x+\cos x}\).

Solution:

Use Weierstrass: \(t = \tan(x/2)\), \(\sin x = \frac{2t}{1+t^2}\), \(\cos x = \frac{1-t^2}{1+t^2}\), \(dx = \frac{2dt}{1+t^2}\).

\int \frac{dx}{1+\sin x+\cos x} = \int \frac{2dt/(1+t^2)}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}

= \int \frac{2dt/(1+t^2)}{\frac{(1+t^2)+2t+(1-t^2)}{1+t^2}} = \int \frac{2dt}{2+2t} = \int \frac{dt}{1+t}

= \ln|1+t| + C = \ln\left|1+\tan\frac{x}{2}\right| + C

\ln\left|1+\tan\frac{x}{2}\right| + C

Answer: \(\ln\left|1+\tan\frac{x}{2}\right| + C\)

17

Reduction Formulae

A reduction formula relates an integral with a higher power to an integral with a lower power. These are derived using integration by parts and are especially useful for integrals of the form \(\int \sin^n x\, dx\), \(\int \cos^n x\, dx\), etc.

17.1 Reduction Formula for \(\int \sin^n x\, dx\)

Most Repeated PYQ — sinⁿ Reduction

\int \sin^n x\, dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2} x\, dx

This reduces the power by 2 each time, until reaching \(\int \sin x\, dx = -\cos x + C\) or \(\int 1\, dx = x + C\).

17.2 Reduction Formula for \(\int \cos^n x\, dx\)

\int \cos^n x\, dx = \frac{\cos^{n-1}x \sin x}{n} + \frac{n-1}{n}\int \cos^{n-2} x\, dx

17.3 Reduction Formula for \(\int \tan^n x\, dx\)

\int \tan^n x\, dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2} x\, dx

Derived using \(\tan^n x = \tan^{n-2}x(\sec^2 x - 1)\).

17.4 Other Reduction Formulae

\int \sec^n x\, dx = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2} x\, dx

\int \csc^n x\, dx = -\frac{\csc^{n-2}x \cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2} x\, dx

\int \sin^m x \cos^n x\, dx = \frac{\sin^{m+1}x \cos^{n-1}x}{m+n} + \frac{n-1}{m+n}\int \sin^m x \cos^{n-2} x\, dx

\int \frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}} + \frac{2n-3}{2a^2(n-1)}\int \frac{dx}{(x^2+a^2)^{n-1}}

17.5 Wallis' Formula (for definite integrals)

Wallis' Formula

\int_0^{\pi/2} \sin^n x\, dx = \int_0^{\pi/2} \cos^n x\, dx = \frac{(n-1)(n-3)\cdots}{n(n-2)\cdots}\times\begin{cases}\frac{\pi}{2} & \text{if } n \text{ even} \\ 1 & \text{if } n \text{ odd}\end{cases}

PYQ — JEE Main 2019 (Jan 11 Shift 1)

Question: Using the reduction formula, evaluate \(\int \sin^4 x\, dx\).

Solution:

Using reduction formula with \(n = 4\):

\int \sin^4 x\, dx = -\frac{\sin^3 x \cos x}{4} + \frac{3}{4}\int \sin^2 x\, dx

Apply again with \(n = 2\):

\int \sin^2 x\, dx = -\frac{\sin x \cos x}{2} + \frac{1}{2}\int 1\, dx = -\frac{\sin x \cos x}{2} + \frac{x}{2}

Substituting back:

\int \sin^4 x\, dx = -\frac{\sin^3 x \cos x}{4} + \frac{3}{4}\left(-\frac{\sin x \cos x}{2} + \frac{x}{2}\right) + C

= -\frac{\sin^3 x \cos x}{4} - \frac{3\sin x \cos x}{8} + \frac{3x}{8} + C

Alternatively (using identities): \(\sin^4 x = \frac{3-4\cos 2x+\cos 4x}{8}\), so:

\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C

Answer: \(\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C\)

18

Important Results & Formulae Summary

A consolidated list of all important results and formulae that frequently appear in JEE problems.

18.1 Complete Formulae Sheet

Most Repeated PYQ — Master Formulae

1. Power: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)

2. Reciprocal: \(\int \frac{1}{x} dx = \ln|x| + C\)

3. Exponential: \(\int e^x dx = e^x + C\), \(\int a^x dx = \frac{a^x}{\ln a} + C\)

4. By Parts: \(\int u\,dv = uv - \int v\,du\)

5. f + f' pattern: \(\int e^x[f(x)+f'(x)]dx = e^x f(x) + C\)

6. 1/(x²+a²): \(\frac{1}{a}\tan^{-1}(x/a) + C\)

7. 1/(x²-a²): \(\frac{1}{2a}\ln|\frac{x-a}{x+a}| + C\)

8. 1/√(a²-x²): \(\sin^{-1}(x/a) + C\)

9. 1/√(x²+a²): \(\ln|x+\sqrt{x^2+a^2}| + C\)

10. 1/√(x²-a²): \(\ln|x+\sqrt{x^2-a^2}| + C\)

11. eˣsin bx: \(\frac{e^x}{1+b^2}(\sin bx - b\cos bx) + C\)

12. eˣcos bx: \(\frac{e^x}{1+b^2}(\cos bx + b\sin bx) + C\)

13. tan x: \(\ln|\sec x| + C\)

14. sec x: \(\ln|\sec x + \tan x| + C\)

15. Weierstrass: \(t = \tan(x/2)\), \(\sin x = \frac{2t}{1+t^2}\), \(\cos x = \frac{1-t^2}{1+t^2}\), \(dx = \frac{2dt}{1+t^2}\)

18.2 Quick Decision Guide

Product of two different function types: Integration by parts (ILATE).
Function and its derivative present: Substitution \(t = f(x)\).
Rational function (P/Q): Partial fractions (if proper) or division + partial fractions (if improper).
1/(ax²+bx+c): Complete the square → arctan or ln form.
1/√(ax²+bx+c): Complete the square → arcsin or ln form.
√(a²-x²): Substitute \(x = a\sin\theta\).
√(a²+x²): Substitute \(x = a\tan\theta\).
√(x²-a²): Substitute \(x = a\sec\theta\).
Rational in sin x, cos x: Weierstrass \(t = \tan(x/2)\).
eˣ × (algebraic + its derivative): Use \(\int e^x[f+f']dx = e^x f + C\).

PYQ — JEE Main 2024 (Jan 29 Shift 1)

Question: The integral \(\int \frac{e^x(1+x)}{\cos^2(xe^x)}dx\) equals:

(a) \(\tan(xe^x) + C\)

(b) \(\sec(xe^x) + C\)

(c) \(\ln|\cos(xe^x)| + C\)

(d) \(-\cot(xe^x) + C\)

Solution:

Let \(t = xe^x\). Then \(dt = (1\cdot e^x + x\cdot e^x)dx = e^x(1+x)dx\).

The integral becomes:

\int \frac{dt}{\cos^2 t} = \int \sec^2 t\, dt = \tan t + C

Substituting back:

\tan(xe^x) + C

Answer: (a) \(\tan(xe^x) + C\)

19

Standard Substitutions Summary

A comprehensive table of standard substitutions for different integrand patterns.

Most Repeated PYQ — Substitution Table

Integrand Contains	Substitution	Result Simplifies To
\(\sqrt{a^2-x^2}\)	\(x = a\sin\theta\)	\(a\cos\theta\)
\(\sqrt{a^2+x^2}\)	\(x = a\tan\theta\)	\(a\sec\theta\)
\(\sqrt{x^2-a^2}\)	\(x = a\sec\theta\)	\(a\tan\theta\)
\(\frac{1}{x^2+a^2}\)	\(x = a\tan\theta\)	\(\frac{1}{a}\tan^{-1}(x/a)\)
\(f(ax+b)\)	\(t = ax+b\)	Standard form
\(\frac{f'(x)}{f(x)}\)	\(t = f(x)\)	\(\ln\|f(x)\|\)
\(f(x)^n \cdot f'(x)\)	\(t = f(x)\)	\(\frac{f(x)^{n+1}}{n+1}\)
Rational in \(\sin x, \cos x\)	\(t = \tan(x/2)\)	Rational in \(t\)
\(\sqrt{\frac{a+x}{a-x}}\) or \(\sqrt{\frac{x}{a-x}}\)	\(x = a\sin^2\theta\) or \(x = a\cos 2\theta\)	Trig integral
\(x = a\cos 2\theta\) form	\(\cos 2\theta = 1-2\sin^2\theta\)	Useful for \(\sqrt{(a-x)/x}\) type

19.1 Special Substitutions for Irrational Functions

\(\sqrt{\frac{a+x}{b+x}}\): substitute \(t^2 = \frac{a+x}{b+x}\).
\(\sqrt{\frac{a-x}{b+x}}\): substitute \(x = a\cos^2\theta + b\sin^2\theta\) or similar.
\((ax+b)^{1/n} \cdot (cx+d)^{1/m}\): substitute \(t = (ax+b)^{1/n}\) if \(n < m\).
For \(\int \frac{dx}{a+b\cos x}\) or \(\int \frac{dx}{a+b\sin x}\): use Weierstrass.

PYQ — JEE Main 2023 (Apr 11 Shift 2)

Question: Evaluate \(\int \frac{dx}{(x^2+1)^2}\) using an appropriate substitution.

Solution:

Use trigonometric substitution: \(x = \tan\theta\), \(dx = \sec^2\theta\, d\theta\).

Then \(x^2 + 1 = \sec^2\theta\), so \((x^2+1)^2 = \sec^4\theta\).

\int \frac{dx}{(x^2+1)^2} = \int \frac{\sec^2\theta\, d\theta}{\sec^4\theta} = \int \cos^2\theta\, d\theta

= \int \frac{1+\cos 2\theta}{2}d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C

Back-substitute: \(\theta = \tan^{-1}x\), \(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2x}{1+x^2}\).

\frac{1}{2}\tan^{-1}x + \frac{x}{2(1+x^2)} + C

Answer: \(\frac{1}{2}\tan^{-1}x + \frac{x}{2(1+x^2)} + C\)

20

Mixed Problems & Special Techniques

Some integrals require creative combinations of the methods we've learned. Here we discuss special techniques and tricky problem types.

20.1 Integration of Rational Functions of \(e^x\)

For \(\int R(e^x)\, dx\), substitute \(t = e^x\), \(dt = e^x\, dx\), so \(dx = dt/t\).

\int R(e^x)\, dx = \int \frac{R(t)}{t}\, dt

20.2 Integration of Hyperbolic Functions

\int \sinh x\, dx = \cosh x + C

\int \cosh x\, dx = \sinh x + C

\int \tanh x\, dx = \ln(\cosh x) + C

\int \text{sech}^2 x\, dx = \tanh x + C

20.3 Special Forms

Useful Special Forms

\int \frac{dx}{a+b\cos x} = \frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\!\left(\frac{(a-b)\tan(x/2)}{\sqrt{a^2-b^2}}\right) + C \quad (a > b)

\int \frac{dx}{a+b\sin x} = \frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\!\left(\frac{a\tan(x/2)+b}{\sqrt{a^2-b^2}}\right) + C \quad (a > b)

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Question: Evaluate \(\int \frac{dx}{e^x + 1}\).

Solution:

Method 1: Substitute \(t = e^x\), \(dt = e^x\, dx\), so \(dx = dt/t\).

\int \frac{dx}{e^x+1} = \int \frac{dt/t}{t+1} = \int \frac{dt}{t(t+1)}

Partial fractions: \(\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}\).

= \ln|t| - \ln|t+1| + C = \ln\left|\frac{t}{t+1}\right| + C = \ln\left(\frac{e^x}{e^x+1}\right) + C

Method 2 (shorter): Multiply numerator and denominator by \(e^{-x}\):

\int \frac{e^{-x}}{1+e^{-x}}dx = -\ln(1+e^{-x}) + C

x - \ln(e^x+1) + C = \ln\left(\frac{e^x}{e^x+1}\right) + C

Answer: \(x - \ln(e^x+1) + C\) (equivalent forms)

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ Integration by parts (ILATE): 20%
✓ Partial fractions: 15%
✓ 1/(ax²+bx+c) type: 14%
✓ Substitution method: 12%
✓ (px+q)/(ax²+bx+c): 10%
✓ Trigonometric integrals: 9%
✓ Special integrals (eˣ forms): 8%
✓ Surd forms 1/√(ax²+bx+c): 7%
✓ Weierstrass substitution: 3%
✓ Reduction formulae: 2%

JEE Advanced (Last 10 Years)

✓ Integration by parts (multi-step): 22%
✓ Special integrals + eˣ forms: 18%
✓ Partial fractions (complex): 15%
✓ Reduction formulae: 12%
✓ Trigonometric substitution: 12%
✓ Euler substitution: 8%
✓ Weierstrass + special forms: 8%
✓ Multi-concept problems: 5%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 4-12 marks (1-3 questions)

Difficulty Level: Easy to Moderate

Time Required: 6-7 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

\(\int (2x^2 - 3x + 1)\, dx\)
\(\int \frac{dx}{x^2+9}\)
\(\int \frac{dx}{\sqrt{16-x^2}}\)
\(\int e^{3x}\, dx\)
\(\int \frac{dx}{2x+3}\)
\(\int \sin^2 x\, dx\)
\(\int x e^x\, dx\)
\(\int \ln x\, dx\)

Level 2: Intermediate (JEE Main/Advanced)

\(\int \frac{x\, dx}{x^2+4x+5}\)
\(\int \frac{dx}{x^2-6x+13}\)
\(\int \frac{dx}{(x-1)(x-2)(x-3)}\)
\(\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)dx\)
\(\int \frac{dx}{1+\sin x}\)
\(\int \sin^3 x \cos^2 x\, dx\)
\(\int \frac{dx}{\sqrt{x^2+4x+3}}\)
\(\int x^2 \sin x\, dx\)

Level 3: Advanced (JEE Advanced/Olympiad)

\(\int \frac{x^2\, dx}{(x^2+1)(x^2+4)}\)
\(\int \frac{dx}{(x^2+1)^2}\) (use reduction formula)
\(\int \frac{\sqrt{x}\, dx}{\sqrt{x}+\sqrt{1-x}}\) (hint: use property \(\int_0^a f(x)dx = \int_0^a f(a-x)dx\))
\(\int e^{2x}\sin 3x\, dx\)
\(\int \frac{dx}{x^4+1}\) (factor and partial fractions)
\(\int \tan^{-1}x\, dx\) (by parts)
Derive the reduction formula for \(\int \sec^n x\, dx\).
\(\int \frac{dx}{a+b\cos x}\) using Weierstrass substitution.

Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is indefinite integration?

Indefinite integration is the reverse process of differentiation. If d/dx[F(x)] = f(x), then ∫f(x)dx = F(x) + C, where C is an arbitrary constant of integration. It is also called anti-derivative.

What is the ILATE rule in integration by parts?

ILATE stands for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. In integration by parts ∫u·dv = u·v - ∫v·du, we choose u as the function that appears first in ILATE order. This helps decide which function to differentiate and which to integrate.

What is the integration by parts formula?

Integration by parts: ∫u·dv = u·v - ∫v·du. It is used when the integrand is a product of two different types of functions. The choice of u and dv follows the ILATE rule.

What are the types of partial fractions?

There are four cases: (1) Distinct linear factors: A/(x-a) + B/(x-b); (2) Repeated linear: A₁/(x-a) + A₂/(x-a)² + ...; (3) Irreducible quadratic: (Ax+B)/(ax²+bx+c); (4) Repeated quadratic: similar with higher powers.

What is the Weierstrass substitution?

The Weierstrass substitution uses t = tan(x/2). Then sin x = 2t/(1+t²), cos x = (1-t²)/(1+t²), and dx = 2dt/(1+t²). It converts any rational function of sin x and cos x into a rational function of t.

How many questions come from Indefinite Integration in JEE?

On average 1-2 questions appear from Indefinite Integration in every JEE Main session. Most repeated topics are: integration by parts, partial fractions, 1/(ax²+bx+c) type, and the eˣ[f(x)+f'(x)] pattern.

What is a reduction formula?

A reduction formula relates an integral with a higher power to an integral with a lower power. For example, ∫sinⁿx dx = (-1/n)sinⁿ⁻¹x·cosx + ((n-1)/n)∫sinⁿ⁻²x dx. These are derived using integration by parts.

Why is there a constant of integration C?

Since the derivative of any constant is zero, the anti-derivative of f(x) is not unique. If F(x) is an anti-derivative, then F(x) + C (for any constant C) is also an anti-derivative. This is why we add C, making the integral "indefinite".

How to choose the right integration method?

Product of two function types → by parts (ILATE). Function and its derivative present → substitution. Rational function → partial fractions. 1/(quadratic) → complete the square. √(a²-x²) → x = a sinθ. √(a²+x²) → x = a tanθ. √(x²-a²) → x = a secθ. Rational in sin/cos → Weierstrass.

What is the golden formula for eˣ integrals?

The golden formula is: ∫eˣ[f(x) + f'(x)]dx = eˣf(x) + C. If you can express the integrand as eˣ times (a function + its derivative), the answer is simply eˣ times that function. This is one of the most asked results in JEE.

Indefinite Integration notes, formulas & PYQs for JEE 2026

📑 Table of Contents

Definition & Basic Concepts

🔥 Most Repeated PYQ — Definition

1.1 Why is it called "Indefinite"?

1.2 Geometrical Interpretation

1.3 Comparison: Differentiation vs Integration

📝 PYQ — JEE Main 2024 (Jan 30 Shift 1)

Standard Integration Formulae

2.1 Power and Algebraic Functions

🔥 Most Repeated PYQ — Power Rule

2.2 Exponential and Logarithmic

2.3 Trigonometric Functions

🔥 Most Repeated PYQ — Trig Integrals

2.4 Inverse Trigonometric Forms

🔥 Most Repeated PYQ — Inverse Trig Forms

2.5 Forms Involving Square Roots

🔥 Most Repeated PYQ — Surd Forms

📝 PYQ — JEE Main 2023 (Apr 10 Shift 2)

Properties of Indefinite Integrals

3.1 Linearity Property

🔥 Most Repeated PYQ — Linearity

3.2 Integral of Derivative

3.3 Derivative of Integral

3.4 Two Integrals are Equal

💡 Equality Condition

3.5 Important Note on Constants

⚠️ Common Mistake

📝 PYQ — JEE Main 2022 (June 28 Shift 1)

Integration by Substitution

4.1 The Substitution Rule

🔥 Most Repeated PYQ — Substitution

4.2 Steps for Substitution

4.3 Standard Substitutions

🔥 Most Repeated PYQ — Standard Substitutions

4.4 Important Special Cases

💡 Must-Remember Results

📝 PYQ — JEE Main 2024 (Apr 5 Shift 2)

Integration of Trigonometric Functions

5.1 Powers of Sine and Cosine

🔥 Most Repeated PYQ — Powers of sin/cos

5.2 Products of Sines and Cosines

🔥 Most Repeated PYQ — Product-to-Sum

5.3 Important Trigonometric Identities for Integration

5.4 Forms Involving tan and cot

💡 Useful Results

📝 PYQ — JEE Main 2024 (Jan 31 Shift 2)

Some Special Integrals

6.1 Exponential-Linear Forms

🔥 Most Repeated PYQ — Exponential Forms

6.2 Special Form: f(x) + f'(x) Pattern

🔥 Most Repeated PYQ — f + f' Pattern

6.3 Logarithmic Forms

6.4 Inverse Trigonometric Functions

📝 PYQ — JEE Main 2023 (Jan 31 Shift 2)

Integration by Parts (ILATE Rule)

7.1 The Formula

🔥 Most Repeated PYQ — Integration by Parts

7.2 The ILATE Rule

🔥 Most Repeated PYQ — ILATE Rule

7.3 Special Forms Using Integration by Parts

7.4 Cyclic Integrals

💡 Cyclic Pattern

📝 PYQ — JEE Main 2024 (Apr 6 Shift 1)

Integration by Partial Fractions

8.1 Proper vs Improper Rational Function

8.2 The Four Cases of Partial Fractions

🔥 Most Repeated PYQ — Partial Fractions Cases

8.3 Heaviside Cover-Up Method (Shortcut)

💡 Shortcut for Case 1

8.4 Integration Results from Partial Fractions

📝 PYQ — JEE Main 2022 (July 25 Shift 1)

Integration of 1/(ax²+bx+c) Form

9.1 Method: Complete the Square

🔥 Most Repeated PYQ — 1/(ax²+bx+c)

9.2 The Three Results

🔥 Most Repeated PYQ — Three Forms

9.3 Summary of Standard Results

📝 PYQ — JEE Main 2021 (Aug 26 Shift 1)

Integration of (px+q)/(ax²+bx+c) Form

Table of Contents

Most Repeated PYQ — Definition

PYQ — JEE Main 2024 (Jan 30 Shift 1)

Most Repeated PYQ — Power Rule

Most Repeated PYQ — Trig Integrals

Most Repeated PYQ — Inverse Trig Forms

Most Repeated PYQ — Surd Forms

PYQ — JEE Main 2023 (Apr 10 Shift 2)

Most Repeated PYQ — Linearity

Equality Condition

Common Mistake

PYQ — JEE Main 2022 (June 28 Shift 1)

Most Repeated PYQ — Substitution

Most Repeated PYQ — Standard Substitutions

Must-Remember Results

PYQ — JEE Main 2024 (Apr 5 Shift 2)

Most Repeated PYQ — Powers of sin/cos

Most Repeated PYQ — Product-to-Sum

Useful Results

PYQ — JEE Main 2024 (Jan 31 Shift 2)

Most Repeated PYQ — Exponential Forms

Most Repeated PYQ — f + f' Pattern

PYQ — JEE Main 2023 (Jan 31 Shift 2)

Most Repeated PYQ — Integration by Parts

Most Repeated PYQ — ILATE Rule

Cyclic Pattern

PYQ — JEE Main 2024 (Apr 6 Shift 1)

Most Repeated PYQ — Partial Fractions Cases

Shortcut for Case 1

PYQ — JEE Main 2022 (July 25 Shift 1)

Most Repeated PYQ — 1/(ax²+bx+c)

Most Repeated PYQ — Three Forms

PYQ — JEE Main 2021 (Aug 26 Shift 1)

Most Repeated PYQ — (px+q)/(ax²+bx+c)

PYQ — JEE Main 2024 (Apr 8 Shift 1)

Most Repeated PYQ — 1/√(ax²+bx+c)

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Most Repeated PYQ — (px+q)/√(ax²+bx+c)

PYQ — JEE Main 2021 (Jul 27 Shift 2)

Most Repeated PYQ — Linear × Quadratic

PYQ — JEE Main 2020 (Sep 5 Shift 1)

Most Repeated PYQ — Trig Substitutions

Additional Substitutions

PYQ — JEE Main 2019 (Apr 10 Shift 1)

Important — Euler's First

Important — Euler's Second

Important — Euler's Third

PYQ — JEE Advanced 2018 (Apr 9 Shift 2)

Most Repeated PYQ — Weierstrass

Usage Guidelines

PYQ — JEE Main 2020 (Oct 5 Shift 1)

Most Repeated PYQ — sinⁿ Reduction

Wallis' Formula

PYQ — JEE Main 2019 (Jan 11 Shift 1)

Most Repeated PYQ — Master Formulae

PYQ — JEE Main 2024 (Jan 29 Shift 1)

Most Repeated PYQ — Substitution Table

PYQ — JEE Main 2023 (Apr 11 Shift 2)

Useful Special Forms

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Previous Year Questions Analysis