Methods of Integration · Ch.7

01

Integration by Substitution

Core Idea

When the integrand contains a composite function, substitute \(t = g(x)\) to reduce it to a standard form. This technique corresponds to the chain rule in reverse.

SUBSTITUTION FORMULA

\[ \int f\bigl(g(x)\bigr)\cdot g'(x)\,dx \;\underset{t=g(x)}{=}\; \int f(t)\,dt \]

Method — Step by Step

1

Choose the substitution

Identify \(t = g(x)\) such that \(g'(x)\,dx\) appears in the integrand.

2

Compute dt

Differentiate: \(dt = g'(x)\,dx\), then express \(dx\) in terms of \(dt\).

3

Replace & integrate

Substitute everything in terms of \(t\) and evaluate \(\int f(t)\,dt\).

4

Back-substitute

Replace \(t\) with the original \(g(x)\) and add \(+C\).

Example — Substitution

Find \(\displaystyle\int \sin(3x+5)\,dx\)

Let \(t = 3x+5 \Rightarrow dt = 3\,dx \Rightarrow dx = \tfrac{dt}{3}\)

\[ \int \sin t\cdot\frac{dt}{3} = -\frac{\cos t}{3}+C = -\frac{\cos(3x+5)}{3}+C \]

Example — Recognising g'(x)

Find \(\displaystyle\int \frac{2x}{x^2+1}\,dx\)

Let \(t = x^2+1 \Rightarrow dt = 2x\,dx\)

\[ \int \frac{dt}{t} = \ln|t|+C = \ln(x^2+1)+C \]

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Useful Pattern: \(\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|+C\). Look for this pattern first!

02

Integration by Parts

Formula — Product Rule in Reverse

Used when the integrand is a product of two different types of functions. Derived from the product rule of differentiation.

IBP FORMULA

\[ \int u\cdot v\,dx = u\int v\,dx \;-\; \int\!\left[\frac{du}{dx}\int v\,dx\right]dx \]

Choosing \(u\) and \(v\): use the ILATE rule to select \(u\) (the one listed first is \(u\)).

ILATE Priority Rule

I

Inverse Trig

L

Logarithmic

A

Algebraic

T

Trigonometric

E

Exponential

The function that appears earlier in ILATE is chosen as \(u\); the other becomes \(v\).

Example 1 — x · eˣ

Find \(\displaystyle\int x\,e^x\,dx\)

By ILATE: \(u = x\) (Algebraic), \(v = e^x\) (Exponential)

\[ = x\cdot e^x - \int 1\cdot e^x\,dx = x e^x - e^x + C = e^x(x-1)+C \]

Example 2 — ln x

Find \(\displaystyle\int \ln x\,dx\)

Write as \(\int \ln x \cdot 1\,dx\). By ILATE: \(u = \ln x\), \(v = 1\)

\[ = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C \]

Special Result — e^x[f(x) + f'(x)]

A beautiful and important result: if the integrand is \(e^x[f(x)+f'(x)]\), we can directly write the answer.

DIRECT FORMULA

\[ \int e^x\bigl[f(x)+f'(x)\bigr]\,dx = e^x\cdot f(x)+C \]

Example: \(\displaystyle\int e^x\!\left(\sin x + \cos x\right)dx = e^x\sin x+C\) (since \((\sin x)' = \cos x\))

03

Integration by Partial Fractions

Core Idea

Any proper rational function \(\dfrac{P(x)}{Q(x)}\) (degree of \(P <\) degree of \(Q\)) can be decomposed into a sum of simpler fractions (partial fractions) that are easy to integrate.

If the fraction is improper (degree of \(P \geq\) degree of \(Q\)), first perform polynomial long division.

Partial Fraction Forms

CASE 1 — LINEAR DISTINCT FACTORS

\[\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}\]

CASE 2 — REPEATED LINEAR FACTOR

\[\frac{px+q}{(x-a)^2} = \frac{A}{x-a}+\frac{B}{(x-a)^2}\]

CASE 3 — IRREDUCIBLE QUADRATIC

\[\frac{px^2+qx+r}{(x-a)(x^2+bx+c)} = \frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}\]

CASE 4 — REPEATED QUADRATIC

\[\frac{P(x)}{(ax^2+bx+c)^2} = \frac{Ax+B}{ax^2+bx+c}+\frac{Cx+D}{(ax^2+bx+c)^2}\]

Example — Partial Fractions

Find \(\displaystyle\int \frac{2x+1}{(x-1)(x+2)}\,dx\)

Decompose: \(\dfrac{2x+1}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}\)

Solving: \(A = 1,\; B = 1\)

\[ \int\!\left(\frac{1}{x-1}+\frac{1}{x+2}\right)\!dx = \ln|x-1|+\ln|x+2|+C = \ln|(x-1)(x+2)|+C \]

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Remember: Always check that the rational function is proper before decomposing. If degree of numerator ≥ degree of denominator, divide first to get: \(\text{quotient} + \text{proper remainder}\).