Can $(x^2+a^2)^{-\frac{3}{2}}$ be integrated without using trigonometric substitutions?

Question

I know that $$ \int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx $$ can be solved using a trigonometric substitution, but is there a trick to calculate this integral without using trigonometric substitutions? I think something in the same spirit with solving $$ \int\frac{1}{(x^2+1)^2}\,dx $$ by adding and subtracting $x^2$ to the numerator and then integrating by parts once one knows (and it is simple) how to integrate $$ \int\frac{1}{(x^2+1)}\,dx $$ Any suggestion or reference will be appreciated.

Answer 1

A very simple approach would be:-

$$\int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx=\int\frac{1}{(x^3)(1+\frac{a^2}{x^2})^{\frac{3}{2}}}\,dx$$

Substituting $(1+\frac{a^2}{x^2})$ as $t^2$,we get :

$$\int\frac{-1}{a^2t^2}dt=\frac{1}{a^2t} +c=\frac{x}{(a^2){(x^2+a^2)^{\frac{1}{2}}}} +c $$

Answer 2

For brevity, let $a=1$

\begin{align} &\int\frac{1}{(x^2+1)^{\frac{3}{2}}}\,dx\\ = &\int \frac{2x^{\frac12}}{(x^2+1)^{\frac14}} d\left( \frac{x^{\frac12}}{(x^2+1)^{\frac14}}\right) = \left( \frac{x^{\frac12}}{(x^2+1)^{\frac14}}\right)^2 = \frac{x}{(x^2+1)^{\frac12}} \end{align}