$\displaystyle \int \frac{sec^2(x)}{1-tan^2(x)}dx$

$\displaystyle u=Tan(x)$

$\displaystyle du=sec^2(x)dx$

$\displaystyle \int \frac{1}{1-u^2}du$

Now, I'm really tempted to use arcsin, but the denominator isn't a radical, so how do I deal with 1/1-u^2 ?