From 0 to 1, integrate x/(x^2+4x+13). I never could understand partial fractions but I have got a few but this one is throwing me off.
$\displaystyle \int\frac{x}{x^2+4x+13}=\int\frac{x+2}{x^2+4x+13}\ ;dx-2\int\frac{dx}{x^2+4x+13}$
complete the square on the second term and make the appropriate substitutions you should end up with this:
$\displaystyle \frac{1}{2}\int\frac{du}{u}-2\int\frac{dx}{(x+2)^2+9}$
second term, you should note, is a standard arctan
Thus finally, $\displaystyle \int\frac{x}{x^2+4x+13}\;dx=\frac{1}{2}\ln(x^2+4x+ 13)-\frac{2}{3}\arctan\left({\frac{x+2}{3}}\right)+C$