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?
 A: The easiest way to solve appears to be to rewrite it as
$$\int\frac{dx}{x^3\sqrt{\frac{1-a^2}{x^2}+1}}$$
Now the substitution 
$$u=(1-a^2)x^{-2}+1,du=2(a^2-1)x^{-3}dx$$ 
reduces the integral to
$$\frac1{2(a^2-1)}\int u^{-1/2}du$$
A: Here's one approach:
$$ \frac{1}{x^2 \sqrt{1-a^2 + x^2}} = \frac{1}{x^3\sqrt{\frac{x^2-a^2+1}{x^2}}}$$
$$ =\frac{1}{x^3\sqrt{\frac{1-a^2}{x^2} + 1}} $$
Let $\theta = \frac{1-a^2}{x^2} + 1$
$$ \frac{d\theta}{dx} = \frac{-2(1-a^2)}{x^3} \ \implies \ dx =  \frac{-x^3}{2(1-a^2)} \ d\theta $$
The integral then reduces to: 
$$ \frac{-1}{2(1-a^2)}\int \frac{1}{\sqrt{\theta}}\ \ d\theta =  \frac{-1}{2(1-a^2)} 2 \sqrt{\theta} + c= -\frac{\sqrt{1-a^2+x^2}}{x(1-a^2)} + c $$
$$ = \frac{\sqrt{1-a^2+x^2}}{x(a^2-1)} + c$$
A: 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
