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: Make the substitution $1+x^2 \to u$ or $x \to \sqrt{u-1}$. Then you can transform the integral into
$$\int -\frac{\sqrt u}{u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$\int -\frac{1}{\sqrt u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$-\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}}du$$
Then complete the square in the denominator:
$$-\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}}du$$
Now see if you can get it in the form
$$\int \frac{1}{\sqrt{a^2-1}}$$
so that you can use the formula
$$\int \frac{1}{\sqrt{a^2-1}}=\ln\bigg(\sqrt{x^2-1}+x\bigg)+C$$
