Partial Fraction Expansion with integral $\int \frac{x^3-5x^2-31x-26}{x(x-2)(x^2+4x+13)} dx$. 
$$\int \frac{x^3-5x^2-31x-26}{x(x-2)(x^2+4x+13)}\,\mathrm dx$$

Hi there, I am having some trouble continuing with this problem.
What I know so far:


*

*That I am able to use partial fraction expansion 

*I'm clear then on how to find A & B, but not C & D

*I don't know if this is because it's such a large problem, but it's really throwing me for a loop. Any guidance would be appreciated.

By $A$ and $B$ I assume you mean that you know how to find $A$ and $B$ where
$$ \dfrac{x^3 - 5 x^2 - 31 x - 26}{x (x-2)(x^2+4x+13)} = \dfrac{A}{x} + \dfrac{B}{x-2} + \dfrac{Cx+D}{x^2 + 4 x + 13}$$
Well, if you've done that correctly, compute
$$ \dfrac{x^3 - 5 x^2 - 31 x - 26}{x (x-2)(x^2+4x+13)} - \dfrac{A}{x} - \dfrac{B}{x-2}$$
with your $A$ and $B$ (putting it over a common denominator).  The $x^2+4x+13$ factor should cancel from numerator and denominator, leaving you with 
$$ \dfrac{Cx+D}{x^2 + 4 x + 13}$$
