Methods of integration

NUST NET (Engineering) · Maths · Integration. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Integration reverses differentiation, so every answer can be checked by differentiating it. Always add the constant c to an indefinite integral.

First look for a standard form. If the integrand contains a function together with its own derivative, use substitution: put u equal to the inside function. If it is a product of two different types, such as x with e^x or x with a trigonometric function, use integration by parts. If it is a fraction whose denominator factorises, split it into partial fractions and integrate each piece as a logarithm.

In multiple choice, differentiating the options is often the fastest method of all.

Rules to remember

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c for n ≠ −1; ∫ (1/x) dx = ln|x| + c
Linear inside: ∫ f(ax + b) dx = (1/a) F(ax + b) + c, e.g. ∫ cos(ax) dx = (1/a) sin(ax) + c
∫ f′(x)/f(x) dx = ln|f(x)| + c; ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + c
∫ dx/(a² + x²) = (1/a) tan⁻¹(x/a) + c; ∫ dx/√(a² − x²) = sin⁻¹(x/a) + c
By parts: ∫ u dv = uv − ∫ v du; choose u in the order logarithm, inverse trig, algebraic, trig, exponential
∫ e^x[f(x) + f′(x)] dx = e^x f(x) + c
∫ dx/(x² − a²) = (1/(2a)) ln|(x − a)/(x + a)| + c; ∫ tan x dx = ln|sec x| + c; ∫ cot x dx = ln|sin x| + c

Common mistake

Multiplying by the inside coefficient instead of dividing, for example writing ∫ e^2x dx = 2e^2x + c. Integration divides: the answer is (1/2) e^2x + c.

Worked example

Find ∫ x e^2x dx.

By parts with u = x and dv = e^2x dx, so du = dx and v = (1/2) e^2x.
∫ x e^2x dx = (x/2) e^2x − ∫ (1/2) e^2x dx = (x/2) e^2x − (1/4) e^2x + c.
Check by differentiating: (1/2) e^2x + x e^2x − (1/2) e^2x = x e^2x.

Answer: (x/2) e^2x − (1/4) e^2x + c

Practice questions

Try each one, then open the answer.

1. ∫ dx/[x(x + 1)] equals

A
ln|(x + 1)/x| + c
B
ln|x(x + 1)| + c
C
ln|x/(x + 1)| + c
D
ln|x| · ln|x + 1| + c

Show answer

Answer: C. Partial fractions: 1/[x(x + 1)] = 1/x − 1/(x + 1), so the integral is ln|x| − ln|x + 1| = ln|x/(x + 1)| + c.

2. ∫ tan² x dx equals

A
tan x − x + c
B
tan x + x + c
C
(tan³ x)/3 + c
D
sec² x + c

Show answer

Answer: A. Use tan² x = sec² x − 1: ∫ (sec² x − 1) dx = tan x − x + c. The power rule does not work here because sec² x is not present as a factor.

3. ∫ ln x dx equals

A
1/x + c
B
x ln x + c
C
x ln x + x + c
D
x ln x − x + c

Show answer

Answer: D. By parts with u = ln x and dv = dx: x ln x − ∫ x × (1/x) dx = x ln x − x + c.

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Methods of integration

The idea

Rules to remember

Common mistake

Worked example

Find ∫ x e2x dx.

Practice questions

1. ∫ dx/[x(x + 1)] equals

2. ∫ tan2 x dx equals

3. ∫ ln x dx equals

Read the full notes

Keep going

Find ∫ x e^2x dx.

2. ∫ tan² x dx equals