Question
Evaluate the following integral:
$\int\frac{1}{\text{a}^2\text{x}^2+\text{b}^2}\text{ dx}$

Answer

$\int\frac{1}{\text{a}^2\text{x}^2+\text{b}^2}\text{ dx}$
$=\frac{1}{\text{a}^2}\int\frac{\text{dx}}{\text{x}^2+\big(\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{1}{\text{a}^2}\times\frac{\text{a}}{\text{b}}\tan^{-1}\bigg(\frac{\text{x}}{\frac{\text{b}}{\text{a}}}\bigg)+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$=\frac{1}{\text{ab}}\tan^{-1}\Big(\frac{\text{ax}}{\text{b}}\Big)+\text{C}$

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