Solving an integral with substitution method I want a hint for the following integral:
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$$
where $a$ is a real constant.
In fact, 
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}} = -\frac{\sqrt{1- a^{2}+x^{2}}}{x(1- a^{2})}$$
why?

Assume a$\ne $$1 or-1$,Put $x$=$\sqrt {1-a^2}$$tan$$\theta$ and then simplify..
$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$=$\int \frac{sec^2\theta d\theta}{({1-a^2})tan^2\theta sec\theta}$=$\int \frac{cos\theta d\theta}{({1-a^2})sin^2\theta}$=
$\int \frac{d(sin\theta)}{({1-a^2})sin^2\theta}$=$\frac{-1}{({1-a^2})sin\theta}$+c
=-$\frac{\sqrt{1-a^2+x^2}}{x(1-a^2)}$+c
