# 6.3 Integration by Substitution

F.5 Mathematics Module 1 · Chapter 6 Indefinite Integration · Lesson 1

Why expanding is not the way, the substitution theorem, the three steps, and the six patterns to spot in an integrand.

## Key points

- **Why substitute**: expanding $(2x + 1)^5$ term by term is tedious, and $\int \frac{1}{(3x - 4)^7}\,dx$ cannot be found from the formulae learnt before. **Integration by substitution ／ 代換積分法** finds both.
- **Theorem**: if $u = g(x)$ is a differentiable function of $x$, then $\int f[g(x)]\,g'(x)\,dx = \int f(u)\,du$. Its proof is the **chain rule ／ 鏈式法則** read backwards.
- **Step 1** substitute $u = g(x)$ and $du = g'(x)\,dx$ to obtain $\int f(u)\,du$. **Step 2** integrate with respect to $u$. **Step 3** replace $u$ by $g(x)$ in the final result.
- **Linear inside** $ax + b$: $u = ax + b$ and $du = a\,dx$, e.g. $\int (9 - 2x)^{11}\,dx = -\tfrac{1}{24}(9 - 2x)^{12} + C$, $\int e^{6x - 9}\,dx = \tfrac{1}{6}e^{6x - 9} + C$.
- **Only a constant is missing?** Multiply and divide by it: $\int x(x^2 + 3)^5\,dx = \tfrac{1}{2}\int (x^2 + 3)^5 \cdot 2x\,dx = (x^2 + 3)^6/12 + C$.
- **Derivative on top**: when the **integrand ／ 被積函數** is $g'(x)/g(x)$, the answer is $\ln\lvert g(x)\rvert + C$, e.g. $\int \frac{9x^2 + 6}{x^3 + 2x - 8}\,dx = 3\ln\lvert x^3 + 2x - 8\rvert + C$.
- **An $x$ left over** (Example 5, $\int \frac{4x}{\sqrt{8x + 7}}\,dx$): put $u = 8x + 7$ and also write $x = (u - 7)/8$, integrate in $u$, then substitute back.
- **Finish the job**: every **indefinite integral ／ 不定積分** ends with $u$ replaced by $g(x)$ and $+\,C$.

## Must know

| The integrand looks like … | Substitute | Answer |
|---|---|---|
| $\int (ax + b)^n\,dx$, $n \ne -1$ | $u = ax + b$, $du = a\,dx$ | $\dfrac{(ax + b)^{n + 1}}{a(n + 1)} + C$ |
| $\int e^{ax + b}\,dx$ | $u = ax + b$, $du = a\,dx$ | $\dfrac{e^{ax + b}}{a} + C$ |
| $\int g'(x)\,[g(x)]^n\,dx$, $n \ne -1$ | $u = g(x)$, $du = g'(x)\,dx$ | $\dfrac{[g(x)]^{n + 1}}{n + 1} + C$ |
| $\int \dfrac{g'(x)}{g(x)}\,dx$ | $u = g(x)$ | $\ln\lvert g(x)\rvert + C$ |
| $\int g'(x)\,e^{g(x)}\,dx$ | $u = g(x)$ | $e^{g(x)} + C$ |
| $\int \dfrac{(\ln x)^n}{x}\,dx$, $n \ne -1$ | $u = \ln x$, $du = \dfrac{1}{x}\,dx$ | $\dfrac{(\ln x)^{n + 1}}{n + 1} + C$ |
| an $x$ is left over, e.g. $\int x\sqrt{2x - 3}\,dx$ | $u = 2x - 3$ and also $x = \dfrac{u + 3}{2}$ | integrate in $u$, then put $u = 2x - 3$ back |

The answer keys write the constant step as $\int (x^2 + 3)^5 \cdot \tfrac{1}{2}\,d(x^2 + 3)$: same working, shorter.

## Common mistakes

- ✗ $\int (9 - 2x)^{11}\,dx = (9 - 2x)^{12}/12 + C$. ✓ $du = -2\,dx$, so divide by $-2$ as well: $-(9 - 2x)^{12}/24 + C$. Leaving out the constant is the commonest slip.
- ✗ Let $u = x^2 + 1$; $\int x(x^2 + 1)^3\,dx = \int u^3\,du = (x^2 + 1)^4/4 + C$. ✓ $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}\,du$ and the answer is $(x^2 + 1)^4/8 + C$. All the $x$ terms and $dx$ must be gone before you integrate in $u$.
- ✗ Stopping at $u^{12}/12 + C$, or forgetting $+\,C$. ✓ Replace $u$ by $g(x)$ in the final answer and write $+\,C$.
- ✗ $\int \frac{g'(x)}{g(x)}\,dx = \ln g(x) + C$. ✓ $\ln\lvert g(x)\rvert + C$. Drop the bars only when $g(x) > 0$ for every $x$, e.g. $e^{2x} + 1$ or $x^2 + 9$.
- ✗ $\int \frac{1}{(3x - 4)^7}\,dx = \tfrac{1}{3}\ln\lvert 3x - 4\rvert + C$. ✓ $\ln$ appears only when the power is 1. Here $u^{-7}$ integrates to $u^{-6}/(-6)$, giving $-\tfrac{1}{18}(3x - 4)^{-6} + C$.

## Quick check

1. Find $\int (2x + 5)^8\,dx$.
2. Find $\int x^2(x^3 + 4)^5\,dx$.
3. Find $\int \dfrac{6x}{x^2 + 4}\,dx$.
4. Find $\int x\,e^{x^2 + 1}\,dx$.
5. Find $\int \dfrac{(\ln x)^2}{x}\,dx$.

## Answers

1. $u = 2x + 5$, $du = 2\,dx$: $\dfrac{(2x + 5)^9}{18} + C$.
2. $u = x^3 + 4$, $du = 3x^2\,dx$: $\dfrac{(x^3 + 4)^6}{18} + C$.
3. $u = x^2 + 4$, $du = 2x\,dx$: $3\ln(x^2 + 4) + C$ (no bars needed: $x^2 + 4 > 0$).
4. $u = x^2 + 1$, $du = 2x\,dx$: $\tfrac{1}{2}e^{x^2 + 1} + C$.
5. $u = \ln x$, $du = \dfrac{1}{x}\,dx$: $\dfrac{(\ln x)^3}{3} + C$.

## Did you know?

Not every integral has a formula: $\int e^{x^2}\,dx$ cannot be written with any combination of the usual functions (polynomials, roots, exponentials, logarithms), a fact proved by Liouville in the 1830s, so no substitution will ever crack it.