Indefinite Integration
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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.