Integration problem $\displaystyle \int \frac{dx}{x(x^3+8)}$ $$\int \frac{dx}{x(x^3+8)}$$
I think I'm supposed to use partial fractions, but I am unsure of how to start the problem. Any help would be appreciated. 

Here's the set up:
$$\int \frac{dx}{x(x^3 + 8)} = \int \frac{dx}{x(x+2)(x^2 -2x + 4)} = \int \left(\frac{A}{x} + \frac{B}{x+2} + \frac{Cx + D}{x^2-2x + 4}\right) \,dx$$
Now you can solve for $A, B, C, D$.
Note: Differences and sums of cubes factor predictably: $$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
In this question, $x^3 + 8 = x^3 + 2^3$.
