Originally Posted by

**joatmon** $\displaystyle \displaystyle \int \frac{x dx}{x^2 + 4x + 13}$

Completing the square,

$\displaystyle \displaystyle \int \frac{x dx}{(x+2)^2 + 9}$

$\displaystyle let u = x + 2; x = u-2; du = dx$

$\displaystyle \int\frac{(u-2)du}{u^2 + 3^2}$

$\displaystyle \int\frac{u du}{u^2 + 3^2} - 2\int\frac{du}{u^2 + 3^2}$

$\displaystyle let v = u^2 + 9; dv = 2u du$

$\displaystyle \frac{1}{2}\int \frac{dv}{v} - \frac{2}{3}\arctan(\frac{u}{3})$

This is actually a definite integral, but I'd like to make sure that I have the antiderivative done correctly, since that is most likely where I have screwed up. Does anybody see any mistakes here?

Thanks!