<s>[INST] 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. 
 [/INST] 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:
$$\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x^2-2x+4}+\frac{Dx}{x^2-2x+4}$$
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