Question
Integrate the rational function in exercise:
$\frac{1}{\text{x}^2-9}$

Answer

$\frac{1}{\text{x}^2-9}\text{dx}$ 
$=\int\frac{1}{\text{x}^2-3^2}\text{dx}$
$=\frac{1}{2\times3}\text{log}\big|\frac{\text{x}-3}{\text{x}+3}\big|+\text{c}$
$\bigg[\therefore \ \int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\text{log}\big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\big|\bigg]$
$=\frac{1}{6}\text{log}\big|\frac{\text{x}-3}{\text{x}+3}\big|+\text{c}$

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