# Integration by Substitution

## Integration by Substitution

Integration by Substitution is performed by replacing the pronumeral (variable) with another pronumeral to simplify the expression for easier integration. Students often made mistakes by forgetting to replace the original pronumeral.

Some functions may be changed to standard forms by easy mathematical manipulation.
To aid the process of changing to a recognisable form, use it also made of substitution, which results in a change of variable. In mathematics, an important use is made of what is called differentials.
\large \begin{align} \displaystyle u &= f(x) \\ du &= \dfrac{du}{dx} \times dx \end{align}

### Question 1

Find $\displaystyle \int{\frac{2x}{\sqrt{x^2-4}}}dx$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{Let } u &= x^2-4 \\ \frac{du}{dx} &= 2x &\color{red} \text{differentiate } u \text{ in terms of } x \\ \frac{du}{2x} &= dx &\color{red} \text{subject to } dx \\ \int{\frac{2x}{\sqrt{x^2-4}}}dx &= \int{\frac{2x}{\sqrt{u}}}\frac{du}{2x} \\ &= \int{\frac{1}{\sqrt{u}}}du \\ &= \int{u^{-\frac{1}{2}}}du \\ &= \dfrac{1}{-\frac{1}{2}+1} u^{\frac{1}{2}} + C \\ &= 2\sqrt{u} + C \\ &= 2\sqrt{x^2-4} + C \end{aligned}

### Question 2

Find $\displaystyle \int{\frac{x^2}{\sqrt{1-x^3}}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= 1-x^3 \\ \frac{du}{dx} &= -3x^2 \\ \frac{du}{-3x^2} &= dx \\ \int{\frac{x^2}{\sqrt{1-x^3}}}dx &= \int{\frac{x^2}{\sqrt{u}}}\frac{du}{-3x^2} \\ &= -\frac{1}{3} \int{\frac{1}{\sqrt{u}}}du \\ &= -\frac{1}{3} \int{u^{-\frac{1}{2}}}du \\ &= -\frac{1}{3} \times \dfrac{1}{-\frac{1}{2}+1} u^{-\frac{1}{2}+1} + C \\ &= -\frac{2}{3} u^{\frac{1}{2}} + C \\ &= -\frac{2}{3} \sqrt{u} + C \\ &= -\frac{2}{3} \sqrt{1-x^3} + C \end{aligned}

### Question 3

Find $\displaystyle \int{x\sqrt{x^2 + 2}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= x^2 + 2 \\ \frac{du}{dx} &= 2x \\ \frac{du}{2x} &= dx \\ \int{x\sqrt{x^2 + 2}}dx &= \int{x\sqrt{u}}\frac{du}{2x} \\ &= \frac{1}{2} \int{\sqrt{u}}du \\ &= \frac{1}{2} \int{u^{\frac{1}{2}}}du \\ &= \frac{1}{2} \times \dfrac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} + C \\ &= \frac{1}{3} u^{\frac{3}{2}} + C \\ &= \frac{1}{3} \sqrt{u^3} + C \\ &= \frac{1}{3} \sqrt{(x^2+2)^3} + C \end{aligned}

### Question 4

Find $\displaystyle \int{\frac{x}{\sqrt{1-x}}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= 1-x \\ \frac{du}{dx} &= -1 \\ dx &= -du \\ u &= 1-x \\ \int{\frac{x}{\sqrt{1-x}}}dx &= \int{\frac{1-u}{\sqrt{u}}}(-du) \\ &= \int{\frac{u-1}{\sqrt{u}}}du \\ &= \int{\frac{u}{\sqrt{u}}}du-\int{\frac{1}{\sqrt{u}}}du \\ &= \int{u^{\frac{1}{2}}}du-\int{u^{-\frac{1}{2}}}du \\ &= \dfrac{1}{\frac{1}{2}+1} u^{\frac{1}{2}+1}-\dfrac{1}{-\frac{1}{2}+1} u^{-\frac{1}{2}+1} + C \\ &= \frac{2}{3} u^{\frac{3}{2}}-2 u^{\frac{1}{2}} + C \\ &= \frac{2}{3} \sqrt{u^3}-2\sqrt{u} + C \\ &= \frac{2}{3} \sqrt{(1-x)^3}-2\sqrt{1-x} + C \end{aligned}

### Question 5

Find $\displaystyle \int{x(x-1)^3}dx$.

\begin{aligned} \displaystyle \text{Let } u &= x-1 \\ \frac{du}{dx} &= 1 \\ dx &= du \\ x &= u+1 \\ \int{x(x-1)^3}dx &= \int{(u+1)u^3}du \\ &= \int{u^4}du + \int{u^3}du \\ &= \frac{1}{5}u^5 + \frac{1}{4}u^4 + C \\ &= \frac{(x-1)^5}{5} + \frac{(x-1)^4}{4} + C \end{aligned}

### Question 6

Find $\displaystyle \int{x^2(x-4)^4}dx$.

\begin{aligned} \displaystyle \text{Let } u &= x-4 \\ \frac{du}{dx} &= 1 \\ dx &= du \\ x &= u+4 \\ \int{x^2(x-4)^4}dx &= \int{(u+4)^2 u^4}dx \\ &= \int{(u^2 + 8u + 16)u^4}du \\ &= \int{u^6}du + 8\int{u^5}du + 16\int{u^4}du \\ &= \frac{1}{7}u^7 + \frac{8}{6}u^6 + \frac{16}{5}u^6 + C \\ &= \frac{1}{7}(x-4)^7 + \frac{4}{3}(x-4)^6 + \frac{16}{5}(x-4)^6 + C \end{aligned}

### Question 7

Find $\displaystyle 2\int{\sqrt{2x+1}}dx$.

\begin{align} \displaystyle \text{Let } u &= 2x+1 \\ \dfrac{du}{dx} &= 2 \\ du &= 2dx \\ 2 \int{\sqrt{2x+1}}dx &= \int{\sqrt{u}}2dx \\ &= \int{u^{\frac{1}{2}}}du \\ &= \dfrac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} +c \\ &= \dfrac{2}{3}u^{\frac{3}{2}} +c \\ &= \dfrac{2\sqrt{u^3}}{3} +c \\ &= \dfrac{2\sqrt{(2x+1)^3}}{3} +c \end{align}

### Question 8

Find $\displaystyle \int{x\sqrt{1+x^2}}dx$.

\begin{align} \displaystyle \text{Let } u &= 1+x^2 \\ \dfrac{du}{dx} &= 2x \\ du &= 2xdx \\ \int{x\sqrt{1+x^2}}dx &= \dfrac{1}{2}\int{\sqrt{1+x^2}}2xdx \\ &= \dfrac{1}{2}\int{\sqrt{u}}du \\ &= \dfrac{1}{2}\int{u^{\frac{1}{2}}}du \\ &= \dfrac{1}{2} \times \dfrac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} +c\\ &= \dfrac{1}{2} \times \dfrac{1}{\frac{3}{2}}u^{\frac{3}{2}} +c \\ &= \dfrac{1}{2} \times \dfrac{2}{3}u^{\frac{3}{2}} +c \\ &= \dfrac{1}{3} \sqrt{u^3} +c \\ &= \dfrac{1}{3} \sqrt{(1+x^2)^3} +c \end{align}

### Question 9

Find $\displaystyle \int{(x^2+3x)^4(2x+3)}dx$.

\begin{align} \displaystyle \text{Let } u &= x^2+3x \\ \dfrac{du}{dx} &= 2x+3 \\ du &= (2x+3)dx \\ \int{(x^2+3x)^4(2x+3)}dx &= \int{u^4}du \\ &= \dfrac{1}{4+1}u^{4+1} +c \\ &= \dfrac{1}{5}u^{5} +c \\ &= \dfrac{1}{5}(x^2+3x)^{5} +c \end{align}