Is this fraction even possible to put into partial fractions? I'm to integrate $\int\frac{x}{x^2+6x+13}dx$
But I'm finding it impossible to do anything with it.
Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...
$\frac{Ax+B}{x^2+6x+13}$ 
Hence $x = (Ax+B)(x^2+6x+13)$
$=>x= Ax^3+(6A+B)x^2+(13A+6B)x+13B$
Equating coefficients mean that A =0, 6A+B = 0, 13A+6B = 1 and 13B = 0.
Obviously this doesn't hold. Since 13A+6B apparently = 1 despite A and B both coming out as zero.
The other way I did it is 
Let x = 0, hence 0 = 13B, therefore B = 0.
Let x = 1, hence 1 = A + 6A + 13A = 20A, therefore A = $\frac{1}{20}$
But that means $\int\frac{x}{x^2+6x+13}dx$ = $\int\frac{x}{20(x^2+6x+13)}dx$, which isn't possible unless x is always 0. And it doesn't help me integrate it.
Quite clearly I've done something wrong something along the lines, but I have no idea what?
Any pointers?

Hints:
$$\frac x{x^2+6x+13}=\frac12\frac{2x+6}{x^2+6x+13}-\frac3{(x+3)^2+4}=$$
$$\frac12\frac{(x^2+6x+13)'}{x^2+6x+13}-\frac32\frac{\left(\frac{x+3}2\right)'}{1+\left(\frac{x+3}2\right)^2}$$
