Antiderivatives, substitution and integration by parts

Goals and prerequisites

Construct antiderivatives and verify them by differentiation. Prerequisites: derivative rules, algebra and domains. This lesson concerns indefinite integration; the next develops definite integrals.

1. Integration reverses differentiation

An antiderivative F of f on an interval satisfies F′=f. All antiderivatives on that interval differ by a constant, because their difference has zero derivative. We write ∫f(x) dx=F(x)+C. The symbol dx identifies the integration variable. Indefinite integration gives a family of functions, not a numerical area.

Linearity gives ∫(af+bg) dx=a∫f dx+b∫g dx for constant a,b. The power rule is ∫xⁿ dx=xⁿ⁺¹/(n+1)+C for n≠−1 on an interval where the powers and derivative rule are valid. The exceptional case is ∫1/x dx=ln|x|+C on any interval avoiding zero. On a disconnected domain, the constants may differ on separate components.

Other basics: ∫eˣ dx=eˣ+C; ∫cos x dx=sin x+C; ∫sin x dx=−cos x+C; ∫1/(1+x²) dx=arctan x+C. Trigonometric derivatives and integrals use radians.

Example: ∫(3x²−4x+2) dx=x³−2x²+2x+C. Differentiating returns the original polynomial, a quick check on coefficients and signs.

2. Substitution is the chain rule reversed

If u=g(x), du=g′(x)dx, and F′=f, then ∫f(g(x))g′(x)dx=F(g(x))+C. The derivative factor must be present or deliberately compensated by a constant. Never change only part of an expression and leave unrelated x terms in a u-integral.

Example: ∫2x/(1+x²) dx. Let u=1+x², du=2x dx. The integral becomes ∫du/u=ln u+C=ln(1+x²)+C; the argument is positive for all real x.

Example: ∫x cos(x²) dx. With u=x² and du=2x dx, x dx=du/2. Hence the answer is (1/2)sin(x²)+C. Differentiation supplies the missing factor 2x and verifies the half.

Example: ∫1/(3x+2) dx=(1/3)ln|3x+2|+C on either interval avoiding −2/3. Omitting 1/3 is a chain-rule error; omitting the domain restriction crosses a singularity.

3. Integration by parts reverses the product rule

From (uv)′=u′v+uv′, ∫u dv=uv−∫v du. Choose u so that differentiation simplifies it, and dv so that an antiderivative is available. This is a strategy rather than a universal mechanical ordering.

For ∫xeˣ dx, take u=x, dv=eˣdx. Then du=dx, v=eˣ, so the result is xeˣ−eˣ+C=(x−1)eˣ+C. Differentiating gives eˣ+(x−1)eˣ=xeˣ.

For ∫ln x dx on x>0, choose u=ln x and dv=dx. Then du=dx/x and v=x, giving x ln x−x+C. For ∫x²eˣ dx, apply parts twice to obtain eˣ(x²−2x+2)+C. The method reduces the polynomial degree each time.

4. Algebraic preparation: partial fractions

Factor a rational denominator and decompose into simpler terms after polynomial division if needed. For 1/[x(x+1)], seek A/x+B/(x+1). Multiplying by x(x+1) gives 1=A(x+1)+Bx, hence A=1 and B=−1. Therefore ∫1/[x(x+1)] dx=ln|x|−ln|x+1|+C, on any interval avoiding −1 and 0.

The decomposition is an algebraic identity only where the original expression is defined. Logs do not remove singularities. Combining the logarithms as ln|x/(x+1)| is allowed on those intervals.

5. Verification and pitfalls

Differentiate every proposed antiderivative. Check the original domain, chain-rule factors, signs and the integration constant. The integral of a product is not the product of the integrals; the integral of a quotient is not the quotient of the integrals. Not every elementary integrand has an elementary antiderivative; a method failing is not proof that an integral is meaningless.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CI01. An antiderivative of 1/x on (−∞,0) is:

A. ln x B. x⁰/0 C. ln|x| D. 1/x²

Answer: C

A: The real logarithm is undefined for negative x.

B: The power rule excludes exponent −1.

C: For x<0, ln|x|=ln(−x) and its derivative is 1/x.

D: Its derivative is −2/x³.

CI02. Evaluate ∫x cos(x²) dx.

A. (1/2)sin(x²)+C B. sin(x²)+C C. −(1/2)sin(x²)+C D. x sin(x²)+C

Answer: A

A: Substitute u=x², so x dx=du/2.

B: Its derivative is twice the integrand.

C: Its derivative has the wrong sign.

D: The product rule adds unwanted terms.

CI03. Evaluate ∫xeˣ dx.

A. xeˣ+C B. (x+1)eˣ+C C. x²eˣ/2+C D. (x−1)eˣ+C

Answer: D

A: Its derivative is (x+1)eˣ.

B: Its derivative is (x+2)eˣ.

C: Its derivative includes an extra x²eˣ/2 term.

D: Integration by parts gives xeˣ−eˣ+C.

CI04. On an interval avoiding −2/3, ∫dx/(3x+2) equals:

A. ln|3x+2|+C B. (1/3)ln|3x+2|+C C. 3ln|3x+2|+C D. −1/(3x+2)²+C

Answer: B

A: Its derivative is three times the integrand.

B: The factor 1/3 cancels the inner derivative 3.

C: Its derivative is nine times the integrand.

D: Its derivative is 6/(3x+2)³.

CI05. For x>0, evaluate ∫ln x dx.

A. 1/x+C B. x ln x+C C. x ln x−x+C D. ln(x²)/2+C

Answer: C

A: This is the derivative of ln x, not its integral.

B: Its derivative is ln x+1.

C: Differentiation gives ln x+1−1.

D: This simplifies to ln x+C and differentiates to 1/x.

CI06. Which decomposition is valid for x≠0,−1?

A. 1/[x(x+1)]=1/x−1/(x+1) B. 1/[x(x+1)]=1/x+1/(x+1) C. 1/[x(x+1)]=x−(x+1) D. 1/[x(x+1)]=1/x²

Answer: A

A: The numerator becomes (x+1)−x=1.

B: Its numerator is 2x+1.

C: The right side is the constant −1.

D: The denominators differ.

Source and scope

First of two lessons interpreting the official Integration heading; elementary antiderivative methods.

Official GATE 2026 CS syllabus (verified 2026-10-14).