Integration by substitution

In a nutshell

Integration by substitution is a more concrete version of the reverse chain rule that makes it easier to integrate a wide variety of functions.

Indefinite integration by substitution

Integration by substitution relies on introducing a new variable, $u(x)$ , and integrating with respect to this variable.

procedure

1.	If not given one, pick a suitable function to write as $u(x)$ .
2.	Differentiate $u$ with respect to $x$ to find an expression for $dx$ .
3.	Rewrite the integral so that only $u$ and $du$ appear.
4.	Evaluate the integral, which will result in a function of $u$ .
5.	Rewrite the answer to give a function of $x$ .

Example 1

Use the substitution $u=x^2+1$ to evaluate $\int \dfrac{7x}{(x^2+1)^2}\, dx$ .

Differentiate $u$ to find $\dfrac{du}{dx}$ and rearrange to obtain an expression for $dx$ :

$\begin{aligned} u&=x^2+1\\\dfrac{du}{dx}&=2x\\du&=2x \, dx\\dx&=\dfrac{1}{2x}\,du \end{aligned}$

Substitute this into the integral, simplifying where applicable:

$\begin{aligned} \int \dfrac{7x}{(x^2+1)^2}\,dx&=\int \dfrac{7\cancel{x}}{(x^2+1)^2}\, \left(\dfrac{1}{2 \cancel{x}}\,du \right)\\&=\int \dfrac{7}{2(x^2+1)^2}\,du \end{aligned}$ 

Use substitution to ensure the integral is only in terms of $u$ :

$\begin{aligned}\int \dfrac{7}{2(x^2+1)^2}\,du &=\int \dfrac{7}{2(u)^2}\,du\\ &=\int \dfrac{7}{2u^2}\,du\end{aligned}$

Perform the integration with $u$ :

$\begin{aligned} \int \dfrac{7}{2u^2}\,du &=\int \dfrac72 u^{-2}\,du \\&=-\dfrac72u^{-1}+C\\&=-\dfrac{7}{2u}+C \end{aligned}$

Use substitution to rewrite the answer in terms of $x$ :

$\begin{aligned} u&=x^2+1\\ -\dfrac{7}{2u}+C&=-\dfrac{7}{2(x^2+1)}+C \end{aligned}$

$\int \dfrac{7x}{(x^2+1)^2}\,dx=\underline{-\dfrac{7}{2(x^2+1)}+C}$

Definite integration by substitution

Using integration by substitution on a definite integral has a very similar process to that of an indefinite integral. The key difference is that the limits also change with the variable.

Example 2

Use the substitution $u=e^x+1$ to evaluate $\int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx$ .

First, differentiate $u$ and find an expression for $dx$ :

$\begin{aligned} u&=e^x+1\\\dfrac{du}{dx}&=e^x \\du&=e^x \, dx\\dx&=\dfrac{1}{e^x}\,du \end{aligned}$

Before substituting, find the corresponding limits for $u$ :

$\space \space u=e^x+1\\ \begin{aligned}x=0&\rightarrow u=e^0+1\rightarrow u=2\\x=\ln(3) &\rightarrow u =e^{\ln(3)}+1\rightarrow u=4 \end{aligned}$

Rewrite the integral in terms of $u$ :

$\begin{aligned} \int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx&=\int_2^4e^{2x}\sqrt{e^x+1}\,\left(\dfrac{1}{e^x}\,du\right)\\&=\int_2^4 e^x\sqrt{e^x+1}\,du\\ &=\int_2^4(u-1)\sqrt{u}\,du\\&=\int_2^4 (u\sqrt u - \sqrt u)\, du\end{aligned}$

Compute the transformed integral:

$\begin{aligned}\int_2^4 (u\sqrt u - \sqrt u)\, du &= \int _2^4 \left(u^\frac32 - u^\frac12\right)\,du\\&=\left[\dfrac25 u^\frac52 - \dfrac23 u^\frac32\right]_2^4\\&=\left(\dfrac25 (4)^\frac52 - \dfrac23 (4)^\frac32\right)-\left(\dfrac25 (2)^\frac52 - \dfrac23 (2)^\frac32\right)\\&=\left(\dfrac25 (32) - \dfrac23 (8)\right) - \left(\dfrac25 (4\sqrt{2}) - \dfrac23 (2\sqrt2)\right)\\&=\dfrac{112}{15}-\dfrac{4}{15}\sqrt2 \end{aligned}$

$\int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx=\underline{\dfrac{4}{15} \left(28-\sqrt2\right)}$