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. 
 A: If you meant $x(x^{3}+8)$ in the denominator of the integral then:
Hint: $$\frac{1}{x(x^{3}+8)}=\frac{1}{8x}-\frac{x^{2}}{8(x^{3}+8)}$$
A: 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, $x^3 + 8 = x^3 + 2^3$.
A: You can let $u=x^3, du=3x^2dx$ to get $\displaystyle\int\frac{1}{x(x^3+8)}dx=\frac{1}{3}\int\frac{3x^2}{x^3(x^3+8)}dx=\frac{1}{3}\int\frac{1}{u(u+8)}du$.  
Now use partial fractions to finish integrating.
A: Hint: Note that $\displaystyle x^3+8=(x+2)(x^2-2x+4)$. It's because you have $a^3+b^3=(a+b)(a^2-ab+b^2)$. For $x^2-2x+4$ you have $\Delta<0$ so you should find partial fractions:
$$\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x^2-2x+4}+\frac{Dx}{x^2-2x+4}$$
A: note that
$$x^5-32 = (x+2)(x^4-2x^3+4x^2-8x+16)$$
$$x^7-128 = (x+2)(x^6-2x^5+4x^4-8x^3+16x^2-32x+64)$$
$$x^9-512 = (x+2)(x^8-2x^7+4x^6-8x^5+16x^4-32x^3+64x^2-128x+256)$$
do you see the pattern?
now apply to pattern to
$$x^3+8$$
