**3.2 Pengamiran Melalui Penggantian**

**1.** Diberi bahawa ${{\displaystyle \int \left(ax+b\right)}}^{n}dx,n\ne -1.$

**(A) Kaedah ****Penggantian****,**

$\begin{array}{l}\text{Katakan}u=ax+b\\ \text{Olehitu,}\frac{du}{dx}=a\\ \text{}\therefore dx=\frac{du}{a}\end{array}$

**Soalan 1:**

$\text{Cari}{{\displaystyle \int \left(3x+5\right)}}^{3}dx.$

*Penyelesaian:*

$\begin{array}{l}\text{Katakan}u=3x+5\\ \text{}\frac{du}{dx}=3\\ \text{}dx=\frac{du}{3}\\ \\ {{\displaystyle \int \left(3x+5\right)}}^{3}dx\\ ={\displaystyle \int {u}^{3}}\frac{du}{3}\text{}\leftarrow \overline{)\begin{array}{l}\text{gantikan}3x+5=u\\ \text{dan}dx=\frac{du}{3}\end{array}}\\ =\frac{1}{3}{\displaystyle \int {u}^{3}}du\\ =\frac{1}{3}\left(\frac{{u}^{4}}{4}\right)+c\\ =\frac{1}{3}\left(\frac{{\left(3x+5\right)}^{4}}{4}\right)+c\text{}\leftarrow \overline{)\begin{array}{l}\text{gantibalik}\\ \text{}u=3x+5\text{}\end{array}}\\ =\frac{{\left(3x+5\right)}^{4}}{12}+c\end{array}$
**(B) Kaedah Rumus**

$\begin{array}{l}{\displaystyle \int {\left(ax+b\right)}^{n}}=\frac{{\left(ax+b\right)}^{n+1}}{\left(n+1\right)a}+c\\ \\ \text{Olehitu,}\\ {\displaystyle \int {\left(3x+5\right)}^{3}dx}=\frac{{\left(3x+5\right)}^{4}}{4\left(3\right)}+c\\ \text{}=\frac{{\left(3x+5\right)}^{4}}{12}+c\end{array}$
**Soalan 2:**

Cari,

$\begin{array}{l}\text{(a)}{\displaystyle \int \frac{2}{7}}{\left(5-x\right)}^{-4}dx\\ \text{(b)}{\displaystyle \int \frac{2}{3{\left(9x-2\right)}^{5}}dx}\end{array}$

*Penyelesaian:*

$\begin{array}{l}\text{(a)}{\displaystyle \int \frac{2}{7}}{\left(5-x\right)}^{-4}dx=\frac{2{\left(5-x\right)}^{-3}}{7\left(-3\right)\left(-1\right)}+c\\ \text{}=\frac{2}{21{\left(5-x\right)}^{3}}+c\end{array}$$\begin{array}{l}\text{(b)}{\displaystyle \int \frac{2}{3{\left(9x-2\right)}^{5}}dx}={\displaystyle \int \frac{2{\left(9x-2\right)}^{-5}}{3}}dx\\ =\frac{2{\left(9x-2\right)}^{-4}}{3\left(-4\right)\left(9\right)}+c\\ =-\frac{2}{108{\left(9x-2\right)}^{4}}+c\\ =-\frac{1}{54{\left(9x-2\right)}^{4}}+c\end{array}$