How to solve this integral easily: $\int \frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$ I am trying to solve this integral $$\int\frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$$
I can do it by brute force (means using a substitution then long division and then substitutions again) but it's too long (suspiciously long solution). Is there any better way to solve it? I guess it should be. If you could provide me with the process that leads to the answer that would really help.

Substitute $\sqrt[3]{x+2}=u$ therefore $\frac{du}{dx} =\dfrac{1}{3\sqrt[3]{(x+2)^2}}$
Your integral will become : 3$\int\dfrac{u^3(u^3-2)}{u^3+u-2}du $
= $3\int [u^3+ \dfrac{5u-2}{4(u^2+u+2)}-u-\dfrac{1}{4(u-1)}]du$
=$3\int u^3du + \dfrac{3}{4}\int(\dfrac{5(2u+1)}{2(u^2+u+2)}-\dfrac{9}{2(u^2+u+2)})du - \dfrac{3}{4}\int\dfrac{1}{u-1}du -3\int u.du$
Now you can  easily see that differentiation of $u^2+u+2 = 2u+1$ so you can substitute $u^2+u+2 = t$ then proceed you will get your result. 
In case of further clarification do let me know...
