<s>[INST] Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$ So I've tried solving the equation below by using $u=x+8$, and I get $\int{\frac{\sqrt[3]{u}}{u-8}}du$ which doesn't seem to lead anywhere, I've also tried taking the $ln$ top and bottom, but I don't know how to proceed. Any hints? 
$$\int{\frac{\sqrt[3]{x+8}}{x}}dx$$ 
Update: Using partial fraction decomposition, $A=2, B=-2, C= -8$ (thank you to all the helpful posts) and after some tedious calculations, the answer I obtained was: $3{\sqrt[3]{x+8}}+24(2ln|\sqrt[3]{x+8}-2|-2ln|\sqrt[3]{x+8}+2|+{\frac{4}{\sqrt[3]{x+8}+2}})+C$
 [/INST] Let $x=u^3-8$, then $\mathrm dx =3u^2\mathrm du$ and we get the integral
$$\int\frac{u}{u^3-8}3u^2\mathrm du$$
Last integral can be solved by partial fractions since
\begin{align*}
\frac{3u^3}{u^3-8}&=3+\frac{24}{u^3-8}\\
&=3+\frac{A}{u-2}+\frac{Bu+C}{u^2+2u+4}
\end{align*}
where $A$, $B$ and $C$ are constants we must find.  
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