Partial Fractions with an irreducible quadratic factor $\int \frac{2}{(x-4)(x^2+2x+6)} dx$. 
this is a partial fraction with irreducible quadratic factors. I know how to set it up and I found A, B, and C. 
2 = A((x^2)+2x+6) +(x-4)(Bx+C). then I plugged in x = 4 to find A. to find the other values I distributed and equated coefficients.
$A = 1/15 \\
B = -1/15 \\
C = -6/15$
I am having trouble with integrating after I plug in the values. Any help is appreciated.

HINT:
Write as
$$\frac2{(x-4)(x^2+2x+6)}=\frac A{x-4}+\frac{B\frac{d(x^2+2x+6)}{dx}+C}{x^2+2x+6}$$
so that $$\int\frac{B\frac{d(x^2+2x+6)}{dx}}{x^2+2x+6}\ dx=B\ln|x^2+2x+6|+C$$
For $\displaystyle\int\frac C{x^2+2x+6}\ dx=\int\frac C{(x+1)^2+(\sqrt5)^2}\ dx$
set $x+1=\sqrt5\tan\theta$
