Now for integrals involving inverse trig functions.

A couple problems:

$\displaystyle \int \frac {1}{1 + 4x^2}dx$

For this one, I know that $\displaystyle \frac {d}{dx} \arctan {x} = \frac {1}{1-x^2}$ but I don't know what to do with the 4 in there.

$\displaystyle \int \frac {1}{x^2 - 6x + 13}dx$

This one I have no clue.

Thanks