**3.7 Pengamiran, SPM Praktis (Kertas 1)**

**Soalan 1:**

*x.*

$\begin{array}{l}(a)\text{}{\displaystyle \int \left(\frac{3}{{x}^{2}}-\frac{5}{2{x}^{3}}+2\right)dx}\\ \\ (b)\text{}{\displaystyle \int {x}^{2}\left({x}^{5}+2x\right)dx}\\ \\ (c)\text{}{\displaystyle \int \frac{3{x}^{4}+2x}{{x}^{3}}dx}\\ \\ (d)\text{}{\displaystyle \int \frac{(7+x)(7-x)}{{x}^{4}}dx}\\ \\ (e)\text{}{\displaystyle \int {(5x-1)}^{3}dx}\\ \\ (f)\text{}{\displaystyle \int \frac{3}{{\left(4x+7\right)}^{8}}dx}\end{array}$

*Penyelesaian:*$\begin{array}{l}\text{(b)}\\ {\displaystyle \int {x}^{2}\left({x}^{5}+2x\right)dx}\\ ={\displaystyle \int \left({x}^{7}+2{x}^{3}\right)}\text{}dx\\ =\frac{{x}^{8}}{8}+\frac{2{x}^{4}}{4}+c\\ =\frac{{x}^{8}}{8}+\frac{{x}^{4}}{2}+c\end{array}$

$\begin{array}{l}\text{(c)}\\ {\displaystyle \int \frac{3{x}^{4}+2x}{{x}^{3}}dx}\\ ={\displaystyle \int \left(\frac{3{x}^{4}}{{x}^{3}}+\frac{2x}{{x}^{3}}\right)}\text{}dx={\displaystyle \int \left(3x+2{x}^{-2}\right)}\text{}dx\\ =\frac{3{x}^{2}}{2}-\frac{2}{x}+c\end{array}$

$\begin{array}{l}\text{(d)}\\ {\displaystyle \int \frac{(7+x)(7-x)}{{x}^{4}}dx}={\displaystyle \int \left(\frac{49-{x}^{2}}{{x}^{4}}\right)}\text{}dx\\ ={\displaystyle \int \left(\frac{49}{{x}^{4}}-\frac{1}{{x}^{2}}\right)\text{}dx}={\displaystyle \int \left(49{x}^{-4}-{x}^{-2}\right)\text{}dx}\\ =\frac{49{x}^{-3}}{-3}+\frac{1}{x}+c\\ =-\frac{49}{3{x}^{3}}+\frac{1}{x}+c\end{array}$

$\begin{array}{l}\text{(e)}\\ {\displaystyle \int {(5x-1)}^{3}dx}\\ =\frac{{(5x-1)}^{4}}{\left(4\right)\left(5\right)}+c\\ =\frac{1}{20}{\left(5x-1\right)}^{4}+c\end{array}$

$\begin{array}{l}\text{(f)}\\ {\displaystyle \int \frac{3}{{\left(4x+7\right)}^{8}}dx}={\displaystyle \int 3{\left(4x+7\right)}^{-8}\text{}dx}\\ =\frac{3{\left(4x+7\right)}^{-7}}{\left(-7\right)\left(4\right)}+c\\ =-\frac{3}{28{\left(4x+7\right)}^{7}}+c\end{array}$