# Indefinite Integration

> Indefinite integration for JEE Mathematics: standard antiderivatives, the a-squared patterns, substitution method and completing the square in quadratics.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/indefinite-integration
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Indefinite Integration", PrepElephant, https://prepelephant.com/topics/jee/mathematics/indefinite-integration

## Direct answer

An antiderivative is a family, not a function: ∫f(x)dx = F(x) + C, every member differing from the next by a constant, and the + C is not decoration — omitting it in JEE costs the answer. The standard patterns are the chapter: ∫dx/(x^2 + a^2) = (1/a)tan^(-1)(x/a) + C; ∫dx/√(a^2 − x^2) = sin^(-1)(x/a) + C; ∫dx/(x^2 − a^2) = (1/2a)ln|(x − a)/(x + a)| + C; ∫dx/√(x^2 + a^2) = ln|x + √(x^2 + a^2)| + C. Before any pattern applies, a quadratic denominator or radicand must be completed to the square — the single most repeated preliminary step in the chapter.

## What you must remember

- **Power and log:** ∫x^n dx = x^(n+1)/(n + 1) + C for n ≠ −1, and ∫dx/x = ln|x| + C — the modulus is mandatory.
- **The a^2 patterns:** x^2 + a^2 leads to arctan; √(a^2 − x^2) leads to sine inverse; x^2 − a^2 leads to the logarithmic split; √(x^2 + a^2) leads to ln|x + √(x^2 + a^2)|.
- **Trigonometric four:** ∫tan x dx = ln|sec x| + C; ∫cot x dx = ln|sin x| + C; ∫sec x dx = ln|sec x + tan x| + C; ∫cosec x dx = ln|cosec x − cot x| + C.
- **Completing the square:** x^2 + 2x + 5 = (x + 1)^2 + 4 — every quadratic must be converted to a shifted standard pattern before matching.
- **Numerator as derivative:** when the numerator is (or hides) the derivative of the denominator, the integral is a logarithm: ∫f'(x)/f(x) dx = ln|f(x)| + C.
- **Substitution:** choose u to absorb the inner function; for indefinite integrals, convert back to x at the end — answers must live in the original variable.
- **Exponential family:** ∫e^x dx = e^x + C and ∫a^x dx = a^x/ln a + C for a > 0, a ≠ 1.

## Two quadratic denominators, done clean

First, ∫dx/(x^2 + 2x + 5). Complete the square: x^2 + 2x + 5 = (x + 1)^2 + 4. The pattern ∫du/(u^2 + a^2) with u = x + 1 and a = 2 gives (1/2)tan^(-1)((x + 1)/2) + C. Second, and more instructive, ∫(2x + 3)/(x^2 + 2x + 5) dx. Split the numerator to expose the derivative of the denominator: 2x + 3 = (2x + 2) + 1. The first piece integrates by the logarithm rule to ln(x^2 + 2x + 5); the second piece is the previous integral, (1/2)tan^(-1)((x + 1)/2). So the answer is ln(x^2 + 2x + 5) + (1/2)tan^(-1)((x + 1)/2) + C. No modulus is needed inside the logarithm because the quadratic is positive for all x (its discriminant is negative). This split — derivative part to the logarithm, remainder to the arctan — resolves a whole genre of JEE Main questions in two moves, and the recognition of which piece is which is the actual skill.

## Where marks leak

JEE Main tests pattern recognition at speed; JEE Advanced forces substitutions before patterns appear — x = a tan θ inside (x^2 + a^2)^(3/2), or u = tan x transforming a rational function of sine and cosine. The recurring losses: matching x^2 − a^2 with the arctan pattern instead of the logarithmic one (the sign inside decides everything); dropping the modulus in ln|...| answers where the argument can be negative; forgetting the constant of integration in questions asking for "the" antiderivative through a point; and leaving an answer in terms of the substitution variable. One silent trap deserves emphasis: after a trigonometric substitution, the back-substitution must restore the original variable correctly — a wrong sign in the right triangle drawn for x = a sin θ flips the final answer's sign.

## Frequently asked questions

### What is ∫dx/(x^2 + a^2)?

(1/a)tan^(-1)(x/a) + C, the pattern behind every completed square with a plus.

### What is ∫dx/√(a^2 − x^2)?

sin^(-1)(x/a) + C, valid for |x| < a.

### Why complete the square before integrating?

Because the standard patterns demand a pure u^2 ± a^2 form; the linear term must be absorbed into the substitution first.

### What is ∫tan x dx?

ln|sec x| + C, with the modulus retained.

### What is ∫dx/(x^2 − a^2)?

(1/2a)ln|(x − a)/(x + a)| + C — the difference of squares always yields a logarithm.