Solve an indefinite integral Find the value of 
$$  \int\frac{\sqrt{1+x^2}}{1-x^2}dx  $$
My Attempt: I tried to arrange the numerator as follows, 
$$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$ 
but that didn't help me.
Any help will be appreciated.

Hint 1: Using $x=\tan(u)$, we get
$$
\begin{align}
\int\frac{\sqrt{1+x^2}}{1-x^2}\,\mathrm{d}x
&=\int\frac{\sec^3(u)}{1-\tan^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec(u)}{\cos^2(u)-\sin^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec^2(u)}{1-2\sin^2(u)}\,\mathrm{d}\sin(u)\\
&=\int\frac{\mathrm{d}\sin(u)}{\left(1-2\sin^2(u)\right)\left(1-\sin^2(u)\right)}\\
\end{align}
$$
Hint 2: Use partial fractions:
$$
\begin{align}
&\frac1{\left(1-2v^2\right)\left(1-v^2\right)}\\
&=\frac2{1-2v^2}-\frac1{1-v^2}\\
&=\frac1{1-\sqrt2v}+\frac1{1+\sqrt2v}-\frac{1/2}{1-v}-\frac{1/2}{1+v}
\end{align}
$$
