Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$

Question

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: $3{\sqrt[3]{x+8}}+24(2ln|\sqrt[3]{x+8}-2|-2ln|\sqrt[3]{x+8}+2|+{\frac{4}{\sqrt[3]{x+8}+2}})+C$

Answer 1

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.

Answer 2

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