<s>[INST] 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.
 [/INST] So, you want to calculate:
$\dfrac{1}{15}\displaystyle{\int}\frac{dx}{x-4}-\dfrac{1}{15}\displaystyle{\int}\frac{x+6}{x^2+2x+6}dx$.
That first integral is just $\ln|x-4|$. For the second one, you want to separate it into two integrals: one natural log and one inverse tangent. Note that the derivative of the denominator is $2x+2=2(x+1)$. You want to see that $x+1$ factor in your numerator, so you rewrite:
$\begin{align}
\displaystyle{\int}\frac{x+6}{x^2+2x+6}dx &= \displaystyle{\int}\frac{x+1+5}{x^2+2x+6}dx \\
&=\displaystyle{\int}\frac{x+1}{x^2+2x+6}dx + \displaystyle{\int}\frac{5}{x^2+2x+6}dx \\
&=\frac12\displaystyle{\int}\frac{2x+2}{x^2+2x+6}dx + \displaystyle{\int}\frac{5}{(x+1)^2+5}dx.
\end{align}$
Note how that last step sets up the integral on the left for substitution with $u=x^2+2x+6$, and sets up the integral on the right for an inverse tangent antiderivative after using the substitution $u=x+1$.
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