Intermediate

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Indefinite Integration

Master Indefinite Integration for JEE Main, JEE Advanced, NEET & Board Exams. Covers all standard formulae for algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions with key identities, the e^x trick and solved examples.

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Integration is the reverse process of differentiation. Given a function $f (x)$ , we seek a function $F (x)$ such that $F^{'} (x) = f (x)$ . The function $F (x)$ is called the antiderivative or primitive of $f (x)$ , and the process is called indefinite integration. Since the derivative of any constant is zero, we always add an arbitrary constant $C$ (the constant of integration) to the result. Mastery of standard formulae and their recognition is the foundation for all higher integration techniques — substitution, by parts, partial fractions — making this the most critical starting point in the entire chapter.

1. Definition and Notation

If $\frac{d}{d x} [F (x)] = f (x)$ , then the indefinite integral of $f (x)$ with respect to $x$ is:

$\int f (x) d x = F (x) + C$

where $C \in R$ is the constant of integration.

$\int$ is the integral sign, $f (x)$ is the integrand, $d x$ indicates the variable of integration.
The indefinite integral represents a family of curves — each value of $C$ gives a different antiderivative, all parallel to each other (differing only by a vertical shift).
Verification: Always verify your answer by differentiating — if $\frac{d}{d x} [F (x) + C] = f (x)$ , the integration is correct.

Two Fundamental Properties

$\frac{d}{d x} [\int f (x) d x] = f (x)$ — differentiation undoes integration.
$\int f^{'} (x) d x = f (x) + C$ — integration undoes differentiation (up to a constant).

2. Standard Formulae — Algebraic and Power Functions

Integrand $f (x)$	$\int f (x) d x$	Condition
$x^{n}$	$\frac{x^{n + 1}}{n + 1} + C$	$n \neq - 1$
$\frac{1}{x}$	$\ln \| x \| + C$	$x \neq 0$
$k$ (constant)	$k x + C$	—
$\frac{1}{\sqrt{x}}$	$2 \sqrt{x} + C$	$x > 0$
$\frac{1}{(a x + b)^{n}}$	$\frac{(a x + b)^{- n + 1}}{a (- n + 1)} + C$	$n \neq 1$
$\frac{1}{a x + b}$	$\frac{1}{a} \ln \| a x + b \| + C$	—
$(a x + b)^{n}$	$\frac{(a x + b)^{n + 1}}{a (n + 1)} + C$	$n \neq - 1$

Golden Rule: For any function $f (a x + b)$ , integrate normally and divide by $a$ (the coefficient of $x$ ). This is the simplest case of the substitution rule.

3. Standard Formulae — Exponential and Logarithmic Functions

Integrand $f (x)$	$\int f (x) d x$
$e^{x}$	$e^{x} + C$
$e^{a x + b}$	$\frac{e^{a x + b}}{a} + C$
$a^{x}$	$\frac{a^{x}}{\ln a} + C$
$a^{a x + b}$	$\frac{a^{a x + b}}{(\ln a) \cdot a} + C$
$\ln x$	$x \ln x - x + C$
$\log_{a} x$	$\frac{x \ln x - x}{\ln a} + C$

4. Standard Formulae — Trigonometric Functions

Integrand $f (x)$	$\int f (x) d x$
$\sin x$	$- \cos x + C$
$\cos x$	$\sin x + C$
$\tan x$	$\ln \| \sec x \| + C = - \ln \| \cos x \| + C$
$\cot x$	$\ln \| \sin x \| + C$
$\sec x$	$\ln \| \sec x + \tan x \| + C$
$\csc x$	$\ln \| \csc x - \cot x \| + C$
$\sec^{2} x$	$\tan x + C$
$\csc^{2} x$	$- \cot x + C$
$\sec x \tan x$	$\sec x + C$
$\csc x \cot x$	$- \csc x + C$
$\sin (a x + b)$	$- \frac{\cos (a x + b)}{a} + C$
$\cos (a x + b)$	$\frac{\sin (a x + b)}{a} + C$

Memory tip: Integrals of co-functions (cos, cot, csc) carry a negative sign — $\int \cos x d x = + \sin x$ is the exception to remember, since $\sin$ is the antiderivative, not a co-function result.

5. Standard Formulae — Inverse Trigonometric Functions

These are among the most frequently tested standard results in JEE. Must be memorised exactly.

Integrand $f (x)$	$\int f (x) d x$
$\frac{1}{\sqrt{1 - x^{2}}}$	$\sin^{- 1} x + C$
$\frac{- 1}{\sqrt{1 - x^{2}}}$	$\cos^{- 1} x + C$
$\frac{1}{1 + x^{2}}$	$\tan^{- 1} x + C$
$\frac{- 1}{1 + x^{2}}$	$\cot^{- 1} x + C$
$\frac{1}{x \sqrt{x^{2} - 1}}$	$\sec^{- 1} x + C$
$\frac{1}{\sqrt{a^{2} - x^{2}}}$	$\sin^{- 1} (\frac{x}{a}) + C$
$\frac{1}{a^{2} + x^{2}}$	$\frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C$
$\frac{1}{x \sqrt{x^{2} - a^{2}}}$	$\frac{1}{a} \sec^{- 1} (\frac{x}{a}) + C$

6. Standard Formulae — Logarithmic (Rational) Results

These results arise from integrating rational expressions and are directly usable as formulae:

Integrand $f (x)$	$\int f (x) d x$
$\frac{1}{x^{2} - a^{2}}$	$\frac{1}{2 a} \ln \| \frac{x - a}{x + a} \| + C$
$\frac{1}{a^{2} - x^{2}}$	$\frac{1}{2 a} \ln \| \frac{a + x}{a - x} \| + C$
$\frac{1}{\sqrt{x^{2} + a^{2}}}$	$\ln \| x + \sqrt{x^{2} + a^{2}} \| + C$
$\frac{1}{\sqrt{x^{2} - a^{2}}}$	$\ln \| x + \sqrt{x^{2} - a^{2}} \| + C$

7. Standard Formulae — Important Integral Forms with Surds

These results are used directly in JEE problems and must be treated as ready-to-use formulae:

Integrand $f (x)$	$\int f (x) d x$
$\sqrt{a^{2} - x^{2}}$	$\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + C$
$\sqrt{x^{2} + a^{2}}$	$\frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \ln \| x + \sqrt{x^{2} + a^{2}} \| + C$
$\sqrt{x^{2} - a^{2}}$	$\frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln \| x + \sqrt{x^{2} - a^{2}} \| + C$

Pattern to notice: All three surd integrals follow the structure:
$\frac{x}{2} \times (surd) \pm \frac{a^{2}}{2} \times (inverse trig or log) + C$

8. Properties of Indefinite Integration

Linearity — Sum/Difference:
$\int [f (x) \pm g (x)] d x = \int f (x) d x \pm \int g (x) d x$
Linearity — Scalar Multiple:
$\int k f (x) d x = k \int f (x) d x$ , where $k$ is a constant.
Two antiderivatives of the same function differ by a constant:
If $F^{'} (x) = G^{'} (x) = f (x)$ , then $F (x) - G (x) = C$ (a constant).
Chain rule in reverse (substitution preview):
$\int f (g (x)) \cdot g^{'} (x) d x = F (g (x)) + C$ where $F^{'} = f$ .

A Crucial Identity

$\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$

This is derived from the product rule: $\frac{d}{d x} [e^{x} f (x)] = e^{x} f (x) + e^{x} f^{'} (x)$ . It is one of the most powerful shortcuts in JEE integration problems. Whenever you spot $e^{x}$ multiplied by a sum of a function and its derivative, the answer is immediately $e^{x} f (x) + C$ .

Extension

$\int e^{g (x)} [f (x) \cdot g^{'} (x) + f^{'} (x)] d x = e^{g (x)} f (x) + C$

9. Key Trigonometric Identities Used in Integration

These identities are used to simplify integrands before integrating. Recognising which identity to apply is a core skill:

Power Reduction (Half-Angle) Identities

$\sin^{2} x = \frac{1 - \cos 2 x}{2}$ $\Rightarrow$ $\int \sin^{2} x d x = \frac{x}{2} - \frac{\sin 2 x}{4} + C$
$\cos^{2} x = \frac{1 + \cos 2 x}{2}$ $\Rightarrow$ $\int \cos^{2} x d x = \frac{x}{2} + \frac{\sin 2 x}{4} + C$
$\tan^{2} x = \sec^{2} x - 1$ $\Rightarrow$ $\int \tan^{2} x d x = \tan x - x + C$
$\cot^{2} x = \csc^{2} x - 1$ $\Rightarrow$ $\int \cot^{2} x d x = - \cot x - x + C$

Product-to-Sum Identities

$\sin A \cos B = \frac{1}{2} [\sin (A + B) + \sin (A - B)]$
$\cos A \cos B = \frac{1}{2} [\cos (A - B) + \cos (A + B)]$
$\sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)]$

These convert products into sums, making integration straightforward using the standard sin/cos formulae.

Pythagorean Identities

$\sin^{2} x + \cos^{2} x = 1$
$1 + \tan^{2} x = \sec^{2} x$
$1 + \cot^{2} x = \csc^{2} x$

10. Quick Recognition Guide

Spotting the Right Formula Instantly

If you see...	Think...	Result involves...
$\frac{1}{\sqrt{a^{2} - x^{2}}}$	Inverse sine	$\sin^{- 1} (x / a)$
$\frac{1}{a^{2} + x^{2}}$	Inverse tangent	$\frac{1}{a} \tan^{- 1} (x / a)$
$\frac{1}{x^{2} - a^{2}}$ or $\frac{1}{a^{2} - x^{2}}$	Partial fractions / log formula	$\ln \| \frac{\cdot}{\cdot} \|$
$\frac{1}{\sqrt{x^{2} \pm a^{2}}}$	Log-type surd formula	$\ln \| x + \sqrt{x^{2} \pm a^{2}} \|$
$\sqrt{a^{2} - x^{2}}$	Surd formula or trig sub $x = a \sin θ$	$\sin^{- 1} (x / a)$ + surd term
$e^{x} [f (x) + f^{'} (x)]$	$e^{x} f (x)$ trick	$e^{x} f (x) + C$
$\sin^{2} x$ or $\cos^{2} x$	Power reduction identity	$\cos 2 x$ terms
$\tan^{2} x$ or $\cot^{2} x$	Pythagorean identity	$\sec^{2} x - 1$ or $\csc^{2} x - 1$
$\frac{f^{'} (x)}{f (x)}$	Logarithm rule	$\ln \| f (x) \| + C$
$f^{'} (x) \cdot [f (x)]^{n}$	Power rule via substitution	$\frac{[f (x)]^{n + 1}}{n + 1} + C$

JEE/NEET Power Tips — Avoid These Common Mistakes

Never forget $+ C$ . In board exams, omitting the constant of integration costs marks every single time. It is not optional — the indefinite integral is a family of functions, not a single function.
$\int \frac{1}{x^{2}} d x \neq \ln | x^{2} |$ . The log rule applies only to $\int \frac{1}{x} d x$ . For $\frac{1}{x^{2}} = x^{- 2}$ , use the power rule: $\int x^{- 2} d x = \frac{x^{- 1}}{- 1} = - \frac{1}{x} + C$ .
$\int \sin^{2} x d x \neq - \frac{\sin^{3} x}{3 \cos x}$ . Always convert using $\sin^{2} x = \frac{1 - \cos 2 x}{2}$ first — do NOT try the power rule on trigonometric functions.
For $\int f (a x + b) d x$ , divide the result by $a$ — the coefficient of $x$ . Students often forget this scaling factor.
$\int e^{f (x)} d x \neq \frac{e^{f (x)}}{f (x)} + C$ in general. The formula $\int e^{a x + b} d x = \frac{e^{a x + b}}{a} + C$ works only for linear exponents.
Two different-looking answers can both be correct. For example, $\int \sin x \cos x d x$ can give $\frac{\sin^{2} x}{2} + C$ or $- \frac{\cos^{2} x}{2} + C$ or $- \frac{\cos 2 x}{4} + C$ — all differ by a constant and are all correct.
Always verify a non-obvious integral by differentiating your answer. This takes 10 seconds and catches errors before they cost marks.

Practice Questions (JEE / NEET / Board Level)

Q1: $\int (x^{3} - \frac{1}{x^{2}} + \sqrt{x} - e^{x} + 3) d x$ equals:

A) $\frac{x^{4}}{4} + \frac{1}{x} + \frac{2}{3} x^{3 / 2} - e^{x} + 3 x + C$
B) $\frac{x^{4}}{4} - \frac{1}{x} + \frac{2 x^{3 / 2}}{3} - e^{x} + 3 x + C$
C) $3 x^{2} + x^{- 3} + \frac{1}{2 \sqrt{x}} - e^{x} + C$
D) $\frac{x^{4}}{4} + \frac{1}{x} + 2 \sqrt{x} - e^{x} + 3 x + C$

Answer: A) $\frac{x^{4}}{4} + \frac{1}{x} + \frac{2}{3} x^{3 / 2} - e^{x} + 3 x + C$ .

Explanation:

Integrate term by term:

$\int x^{3} d x = \frac{x^{4}}{4}$

$\int - x^{- 2} d x = - \frac{x^{- 1}}{- 1} = \frac{1}{x}$

$\int x^{1 / 2} d x = \frac{x^{3 / 2}}{3 / 2} = \frac{2}{3} x^{3 / 2}$

$\int - e^{x} d x = - e^{x}$

$\int 3 d x = 3 x$

Combined: $\frac{x^{4}}{4} + \frac{1}{x} + \frac{2}{3} x^{3 / 2} - e^{x} + 3 x + C$ .

Q2: $\int \frac{d x}{9 + 4 x^{2}}$ equals:

A) $\frac{1}{6} \tan^{- 1} (\frac{2 x}{3}) + C$
B) $\frac{1}{3} \tan^{- 1} (\frac{2 x}{3}) + C$
C) $\frac{1}{12} \tan^{- 1} (\frac{2 x}{3}) + C$
D) $\tan^{- 1} (2 x) + C$

Answer: A) $\frac{1}{6} \tan^{- 1} (\frac{2 x}{3}) + C$ .

Explanation:

Factor out 4 to match the standard integral form:

Write $9 + 4 x^{2} = 4 (\frac{9}{4} + x^{2}) = 4 [{(\frac{3}{2})}^{2} + x^{2}]$

Substitute into the integral:
$\int \frac{d x}{9 + 4 x^{2}} = \frac{1}{4} \int \frac{d x}{{(\frac{3}{2})}^{2} + x^{2}}$

Evaluate using the standard form:
$= \frac{1}{4} \cdot \frac{1}{3 / 2} \tan^{- 1} (\frac{x}{3 / 2}) + C$
$= \frac{1}{4} \cdot \frac{2}{3} \tan^{- 1} (\frac{2 x}{3}) + C$
$= \frac{1}{6} \tan^{- 1} (\frac{2 x}{3}) + C$ .

Q3: $\int \cos^{2} x d x$ equals:

A) $\frac{\sin 2 x}{4} + C$
B) $\frac{x}{2} + \frac{\sin 2 x}{4} + C$
C) $\frac{x}{2} - \frac{\sin 2 x}{4} + C$
D) $\frac{\cos^{3} x}{3} + C$

Answer: B) $\frac{x}{2} + \frac{\sin 2 x}{4} + C$ .

Explanation:

Use the half-angle identity $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ :

$\int \cos^{2} x d x = \int \frac{1 + \cos 2 x}{2} d x$
$= \frac{1}{2} \int 1 d x + \frac{1}{2} \int \cos 2 x d x$
$= \frac{x}{2} + \frac{1}{2} \cdot \frac{\sin 2 x}{2} + C$
$= \frac{x}{2} + \frac{\sin 2 x}{4} + C$ .

Q4: $\int e^{x} (\sin x + \cos x) d x$ equals:

A) $e^{x} \cos x + C$
B) $e^{x} \sin x + C$
C) $e^{x} (\sin x - \cos x) + C$
D) $2 e^{x} \sin x + C$

Answer: B) $e^{x} \sin x + C$ .

Explanation:

Recognise the standard form $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ .

Here, $f (x) = \sin x$ and its derivative is $f^{'} (x) = \cos x$ .

Therefore, $\int e^{x} (\sin x + \cos x) d x = e^{x} \sin x + C$ .

Q5: $\int \frac{1}{\sqrt{7 - 6 x - x^{2}}} d x$ equals:

A) $\sin^{- 1} (\frac{x + 3}{4}) + C$
B) $\sin^{- 1} (\frac{x - 3}{4}) + C$
C) $\sin^{- 1} (\frac{x + 3}{3}) + C$
D) $- \cos^{- 1} (\frac{x + 3}{4}) + C$

Answer: A) $\sin^{- 1} (\frac{x + 3}{4}) + C$ .

Explanation:

Complete the square in the expression under the radical:

$7 - 6 x - x^{2} = - (x^{2} + 6 x - 7)$
$= - [(x + 3)^{2} - 9 - 7]$
$= - [(x + 3)^{2} - 16]$
$= 16 - (x + 3)^{2}$

Substitute this back into the integral:
$\int \frac{d x}{\sqrt{16 - (x + 3)^{2}}}$

Using the standard form $\int \frac{d x}{\sqrt{a^{2} - u^{2}}} = \sin^{- 1} (\frac{u}{a}) + C$ with $u = x + 3$ and $a = 4$ :
$= \sin^{- 1} (\frac{x + 3}{4}) + C$ .

Q6: $\int \tan^{2} x d x$ equals:

A) $\sec^{2} x + C$
B) $\tan x + x + C$
C) $\tan x - x + C$
D) $2 \tan x - x + C$

Answer: C) $\tan x - x + C$ .

Explanation:

Use the trigonometric identity $\tan^{2} x = \sec^{2} x - 1$ :

$\int \tan^{2} x d x = \int (\sec^{2} x - 1) d x$
$= \tan x - x + C$ .

Q7 (JEE Advanced type): $\int \frac{x^{2} + 1}{x^{4} + 1} d x$ . Which substitution simplifies this most efficiently?

A) $x = \tan θ$
B) Divide numerator and denominator by $x^{2}$ , then substitute $t = x - \frac{1}{x}$
C) $x^{2} = t$
D) Partial fractions directly on $\frac{x^{2} + 1}{x^{4} + 1}$

Answer: B) Divide by $x^{2}$ , then substitute $t = x - \frac{1}{x}$ .

Explanation:

Divide both the numerator and denominator by $x^{2}$ :

$\frac{x^{2} + 1}{x^{4} + 1} = \frac{1 + 1 / x^{2}}{x^{2} + 1 / x^{2}}$

Complete the square in the denominator:
$x^{2} + \frac{1}{x^{2}} = {(x - \frac{1}{x})}^{2} + 2$

Let $t = x - \frac{1}{x}$ , which means $d t = (1 + \frac{1}{x^{2}}) d x$ . This perfectly matches the numerator factor.

Substitute into the integral:
$\int \frac{1 + 1 / x^{2}}{(x - 1 / x)^{2} + 2} d x = \int \frac{d t}{t^{2} + 2}$

Evaluate the integral:
$= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t}{\sqrt{2}}) + C$
$= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x^{2} - 1}{x \sqrt{2}}) + C$ .

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Indefinite Integration

1. Definition and Notation

Two Fundamental Properties

2. Standard Formulae — Algebraic and Power Functions

3. Standard Formulae — Exponential and Logarithmic Functions

4. Standard Formulae — Trigonometric Functions

5. Standard Formulae — Inverse Trigonometric Functions

6. Standard Formulae — Logarithmic (Rational) Results

7. Standard Formulae — Important Integral Forms with Surds

8. Properties of Indefinite Integration

A Crucial Identity

Extension

9. Key Trigonometric Identities Used in Integration

Power Reduction (Half-Angle) Identities

Product-to-Sum Identities

Pythagorean Identities

10. Quick Recognition Guide

Spotting the Right Formula Instantly

Practice Questions (JEE / NEET / Board Level)

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Practice Questions

Topic Information

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