Question:
Evaluate the following integrals:
$\int \frac{1}{a^{2} x^{2}+b^{2}} d x$
Solution:
take out $a^{2}$
$=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\frac{b^{2}}{a^{2}}} d x$
$=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\left(\frac{b}{a}\right)^{2}} d x=\frac{1}{a^{2}} * \frac{1}{\left(\frac{b}{a}\right)} \tan ^{-1}\left[\frac{x}{\frac{b}{a}}\right]+c\left\{\right.$ since $\left.\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{b}{a}\right)+c\right\}$
$=\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\mathrm{ax}}{\mathrm{b}}\right)+\mathrm{c}$
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