Let $I=\int \frac{d x}{\left(x^2-a^2\right)^{3 / 2}}$
Putting $x=a \sec \theta \Rightarrow d x=a \sec \theta \cdot \tan \theta d \theta$
$$
\begin{aligned}
& I=\int \frac{a \sec \theta \cdot \tan \theta \cdot d \theta}{\left[a^2\left(\sec ^2 \theta-1\right)\right]^{\frac{3}{2}}}=\int \frac{1}{a^2} \times \frac{\sec \theta \cdot \tan \theta}{\tan ^3 \theta^2} d \theta \\
&=\frac{1}{a^2} \int \frac{1}{\cos \theta \cdot \frac{\sin ^2 \theta}{\cos ^2 \theta}} d \theta=\frac{1}{a^2} \int \frac{\cos \theta}{\sin ^2 \theta} d \theta \\
&=\frac{1}{a^2} \int \cot \theta \cdot \operatorname{cosec} \theta d \theta=-\frac{1}{a^2} \operatorname{cosec}^2 \theta+C \\
&=-\frac{x}{a^2 \sqrt{x^2-a^2}}+C \\
&\left(\because \sec \theta=\frac{x}{a} \Rightarrow \cos \theta=\frac{a}{x} \therefore \sin \theta=\sqrt{1-\frac{a^2}{x^2}}\right. \\
&\left.\quad \Rightarrow \cos \theta=\sqrt{\frac{x^2}{x^2-a^2}}=\frac{x}{\sqrt{x^2-a^2}}\right)
\end{aligned}
$$