JEE Mains Maths · Indefinite Integration

Algebraic Substitution

Choosing a substitution that clears a root, a pair of linear factors or a high power of x, so the integral becomes a power of t.

All Indefinite Integration notes JEE Mains Maths strategy

Why this matters

Fourteen PYQs, eight of them multiple choice, and two from 2026. Seven clear a root, with x = tⁿ, with t² equal to the expression under the root, with t = √(1 + x²) + x, or with x = cos θ; four have two linear factors whose powers add to 2, settled by their ratio; three take the leading power of x out of a bracket, or put x = 1/t. Three ideas cover the page.

Concept 1 of 3: Clearing a root

A root blocks every standard formula, so the first move is a substitution that removes it. Several fractional powers of

x

clear together with

x=t^n

, where

n

is the least common multiple of the root orders. A root of an expression clears when

t^2

equals that expression. After the change, the integrand is a polynomial or a simple fraction in

t

Definition

Powers $x^{p/q}$ of different orders: put $x=t^n$ , $n$ the LCM of the denominators.
$\sqrt{a\pm x^2}$ next to an odd power of $x$ , or $\sqrt{ax+b}$ : put $t^2$ equal to the expression.
$\sqrt{1+x^2}\pm x$ : put $t=\sqrt{1+x^2}+x$ ; then $\sqrt{1+x^2}-x=\frac1t$ .
$\sqrt{\frac{1-x}{1+x}}$ : put $x=\cos\theta$ , and the root becomes $\tan\frac\theta2$ .

Mixed roots of x

x=t^{n},\quad dx=nt^{n-1}\,dt,\quad n=\operatorname{lcm}(\text{root orders})

Worked example

Find

\int x^3\sqrt{1+x^2}\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q69Moderate

Example 1 · Indefinite Integration · Algebraic Substitution

Let

f(x) =\int_{}^{}\ \frac{dx}{x^{\left( \frac{2}{3} \right)}+ 2x^{\left( \frac{1}{2} \right)}}

be such that

f(0) = - 26 + 24\log_{e}(2)

. If

f(1) =a+b\log_{e}(3)

, where

a,b\in Z

, then

a + b

is equal to:

Given values are in x

With

x=t^6

, the point

x=64

t=2

, not

t=64

. Convert the point to

t

, or the answer back to

x

, before fixing the constant.

Drill 6 more on clearing a root

Concept 2 of 3: Two linear factors: use their ratio

When the integrand is one over two linear factors raised to powers that add to 2, divide by the square of one factor. What is left is a power of their ratio, and the derivative of that ratio is a constant over the same square. So the ratio is the substitution, and the integral is a single power of

t

Definition

For $\frac{1}{(x-a)^p(x+b)^q}$ with $p+q=2$ , write it as $\frac{1}{t^p(x+b)^2}$ , where $t=\frac{x-a}{x+b}$ .
$dt=\frac{(a+b)\,dx}{(x+b)^2}$ , so the integral is $\frac{1}{a+b}\int t^{-p}\,dt$ .
A linear factor times the root of a quadratic: factor the quadratic first. The powers often add to 2.

Ratio substitution

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

Worked example

Find

\int\frac{dx}{\sqrt[3]{(x-1)^2(x+2)^4}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q64Moderate

Example 2 · Indefinite Integration · Algebraic Substitution

Let

I(x) = \int\frac{3dx}{(4x + 6)\left( \sqrt{4x^{2}+ 8x + 3} \right)}

and

I(0) =\frac{\sqrt{3}}{4}+ 20

. If

I\left( \frac{1}{2} \right)=\frac{a\sqrt{2}}{\text{ }b}+ c

, where

a,b,c \in N,gcd(a,b) = 1

, then

a + b + c

is equal to :

Differentiate the ratio in full

For

t=\frac{2x+1}{2x+3}

dt=\frac{4\,dx}{(2x+3)^2}

, not

\frac{2\,dx}{(2x+3)^2}

. The constant on top is

2\cdot3-2\cdot1

, and a wrong constant scales the whole answer.

Drill 3 more on two linear factors: use their ratio

Concept 3 of 3: Taking out a power of x

A bracket of high powers of

x

often hides a simple function of

\frac1x

. Take the leading power out of the bracket, or divide top and bottom by a power of

x

, and the rest of the integrand becomes the derivative of what is inside. Then the integral is a power of the bracket. For a quadratic inside a root beside another quadratic,

x=\frac1t

does the same job.

Definition

$(x^m+x^n)^k$ : take the leading power out, so the bracket becomes a function of $\frac1x$ .
$\frac{P(x)}{Q(x)^2}$ with high powers: divide top and bottom by a power of $x$ until the top is the derivative of the new bracket.
$\frac{1}{(ax^2+b)\sqrt{cx^2+d}}$ : put $x=\frac1t$ , then $z^2$ equal to the root.
Finish with $\int f'f^k\,dx=\frac{f^{k+1}}{k+1}$ .

Power rule for a bracket

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

Worked example

Find

\int\frac{2x^3+3x^2}{(x^3+x+1)^2}\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 7 Apr 2025 · Q147Moderate

Example 3 · Indefinite Integration · Algebraic Substitution

\int\left( \frac{1}{x} + \frac{1}{x^{3}} \right)\left( \sqrt[23]{3x^{- 24} + x^{- 26}} \right)dx = - \frac{\alpha}{3(\alpha + 1)}\left( 3x^{\beta} + x^{\gamma} \right)^{\frac{\alpha + 1}{\alpha}} + C,\ x > 0

(\alpha,\beta,\gamma \in Z)

, where

C

is the constant of integration, then

\alpha + \beta + \gamma

is equal to .

A root of a power

\sqrt[n]{x^mg}=x^{m/n}\sqrt[n]{g}

needs

x>0

when

n

is even; otherwise

\sqrt{x^2}=|x|

and a sign appears. The condition

x>0

in the question is what allows the step.

Drill 2 more on taking out a power of x

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Clearing a root
Mixed roots of x
$x=t^{n},\quad dx=nt^{n-1}\,dt,\quad n=\operatorname{lcm}(\text{root orders})$
Two linear factors: use their ratio
Ratio substitution
$\int\frac{dx}{(x-a)^p(x+b)^{2-p}}=\frac{1}{(a+b)(1-p)}\left(\frac{x-a}{x+b}\right)^{1-p}+C$
Taking out a power of x
Power rule for a bracket
$\int f'(x)\,[f(x)]^{k}\,dx=\frac{[f(x)]^{k+1}}{k+1}+C,\quad k\neq-1$

Watch out for (3)

Given values are in x
→ Clearing a root
Differentiate the ratio in full
→ Two linear factors: use their ratio
A root of a power
→ Taking out a power of x

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