Answer
$$\displaystyle{\int\frac{1}{\left(\left(ax\right)^2-b^2\right)^\frac{3}{2}}dx=-\frac{x}{b^2\sqrt{\left(ax\right)^2-b^2}}+C}\\$$
Work Step by Step
$\displaystyle{I=\int\frac{1}{\left(\left(ax\right)^2-b^2\right)^\frac{3}{2}}dx}\\$
$\displaystyle{I=\int\frac{1}{\left(b^2\sec^2x-b^2\right)^\frac{3}{2}}\times\frac{b}{a}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{1}{\left(b^2\left(\sec^2x-1\right) \right)^\frac{3}{2}}\times\frac{b}{a}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int\frac{\sec\theta}{ab^2\tan^2\theta}}\\ \displaystyle{I=\frac{1}{ab^2}\int\frac{1}{\cos\theta}\times\frac{\cos^2\theta}{\sin^2\theta} d\theta}\\ \displaystyle{I=\frac{1}{ab^2}\int\cot\theta\times cosec\theta\ d\theta}\\ \displaystyle{I=\frac{1}{ab^2}\left(-\cot\theta\right)+C}\\ \displaystyle{I=-\frac{1}{ab^2}\cot\theta+C}\\$
$\cot\theta=\frac{ax}{\sqrt{\left(ax\right)^2-b^2}}\\$
$\displaystyle{I=-\frac{1}{ab^2}\times\frac{ax}{\sqrt{\left(ax\right)^2-b^2}}+C}\\ \displaystyle{I=-\frac{x}{b^2\sqrt{\left(ax\right)^2-b^2}}+C}\\$
