Find $\displaystyle\int 2x(x^2 + 1)^3\,dx$.

$\dfrac{x^2(x^2 + 1)^4}{4} + C$

Integration by Substitution | HSC Maths Extension 1

In short: Integration by substitution works when a function and the derivative of its inside both appear. Choose $u$ for the inside, replace $d x$ using $d u$ , integrate in $u$ , then change back to $x$ . For a definite integral, change the limits to $u$ values too.

Starting point: what you already know

The chain rule: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$ . For example $\frac{d}{d x} (x^{2} + 1)^{6} = 6 (x^{2} + 1)^{5} \cdot 2 x$ .
Basic integrals: $\int u^{n} d u = \frac{u ^{n + 1}}{n + 1} + C$ for $n \neq = - 1$ , $\int e^{u} d u = e^{u} + C$ , $\int \frac{1}{u} d u = ln ∣ u ∣ + C$ , $\int cos u d u = sin u + C$ .
Definite integrals: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .
Index laws: $u = u^{1/2}$ and $\frac{1}{u} = u^{- 1/2}$ .

The idea

Look at $\int 2 x (x^{2} + 1)^{5} d x$ . The inside, $x^{2} + 1$ , has derivative $2 x$ , and $2 x$ is sitting right there. That is the sign that substitution will work.

Definite integrals: change the limits

When $x$ goes from $a$ to $b$ , work out where $u$ goes. Then you never have to switch back to $x$ .

Integration by substitution

When to use

Inside function
Its derivative is also there

Steps

Choose u
Find du
Replace x and dx
Integrate in u
Back to x

Definite integrals

Change the limits
No need to go back to x

Check

Differentiate your answer

Year 11The exam gives you the substitution

Use $u = x + 1$ to find $\int_{0}^{3} \frac{x}{x + 1} d x$ .

Here $d u = d x$ and $x = u - 1$ . Limits: $x = 0$ gives $u = 1$ , $x = 3$ gives $u = 4$ .

$\int_{1}^{4} \frac{u - 1}{u} d u = \int_{1}^{4} (u^{1/2} - u^{- 1/2}) d u = [\frac{2}{3} u^{3/2} - 2 u^{1/2}]_{1}^{4}$

$= (\frac{16}{3} - 4) - (\frac{2}{3} - 2) = \frac{8}{3}$ .

The trick: write the leftover $x$ in terms of $u$ too.

When the limits flip. With $u = 1 - x$ , $d u = - d x$ , and $x = 0, 1$ become $u = 1, 0$ . The minus sign flips them back: $\int_{0}^{1} x (1 - x)^{4} d x = \int_{0}^{1} (1 - u) u^{4} d u = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}$ .

A squared substitution. With $u^{2} = x + 1$ : $x = u^{2} - 1$ and $d x = 2 u d u$ . This removes the square root completely: $\frac{x}{x + 1} d x$ becomes $\frac{u ^{2} - 1}{u} \cdot 2 u d u = 2 (u^{2} - 1) d u$ . Similarly $u = x$ means $x = u^{2}$ and $d x = 2 u d u$ , so $\int_{0}^{4} \frac{d x}{1 + x} = \int_{0}^{2} \frac{2 u}{1 + u} d u = 2 \int_{0}^{2} (1 - \frac{1}{1 + u}) d u = 4 - 2 ln 3$ .

Logs. With $u = ln x$ , $d u = \frac{d x}{x}$ : $\int \frac{ln x}{x} d x = \frac{( ln x ) ^{2}}{2} + C$ and $\int \frac{d x}{x ln x} = ln ∣ ln x ∣ + C$ .

Quick summary

Use substitution when an inside function and its derivative both appear.
Choose $u$ , find $d u$ , replace everything, integrate, then go back to $x$ .
For a definite integral, change the limits to $u$ values.
Fix missing constants: if $d u = 3 x^{2} d x$ then $x^{2} d x = \frac{1}{3} d u$ .
Differentiate your answer to check it.

Quiz

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Try 3 questions now

For $\int 3 x^{2} (x^{3} - 2)^{4} d x$ , the best choice of $u$ is:
1. A $u = 3 x^{2}$
2. B $u = x^{3} - 2$
3. C $u = (x^{3} - 2)^{4}$
4. D $u = x$
Show answer
B: $u = x^{3} - 2$
The inside is $x^{3} - 2$ and its derivative $3 x^{2}$ is present.
If $u = x^{2} + 5$ , then $d u$ equals:
1. A $x d x$
2. B $2 x d x$
3. C $(2 x + 5) d x$
4. D $x^{2} d x$
Show answer
B: $2 x d x$
$\frac{d u}{d x} = 2 x$ .
Find $\int 2 x (x^{2} + 1)^{3} d x$ .
1. A $\frac{( x ^{2} + 1 ) ^{4}}{4} + C$
2. B $(x^{2} + 1)^{4} + C$
3. C $\frac{x ^{2} ( x ^{2} + 1 ) ^{4}}{4} + C$
4. D $\frac{( x ^{2} + 1 ) ^{4}}{8} + C$
Show answer
A: $\frac{( x ^{2} + 1 ) ^{4}}{4} + C$
With $u = x^{2} + 1$ it becomes $\int u^{3} d u$ .

Integration by Substitution