<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] you can start with
$$\begin{align}
u^3=x+8&\Rightarrow 3u^2du=dx\\
x=u^3-8\\
\int\frac{\sqrt[3]{x+8}}{x}dx&=\int\frac{\sqrt[3]{u^3}}{u^3-8}(3u^2du)\\
&=\int\frac{3u^3}{u^3-8}\\
&=\int\frac{3(u^3-8)+24}{u^3-8}du\\
&=\int3+\frac{24}{u^3-8}du
\end{align}$$
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