Integral with roots on bottom and top

[rasen58]

It's the integral of sqrt(x)/(cubed root(x) + 1)
I tried regular u substitution but that didn't let me get rid of all the x's.
I also just tried long division but that gave me an answer that didn't match with the actual answer to the problem.

The actual answer is 6[1/7 x^(7/6) - 1/5 x^(5/6) + 1/3 x^(1/6) - x^(1/6) + arctan(x^(1/6)) ] + C
How do I go about doing this?

 

[mfb]

The roots suggest a substitution of x^1/6 as this allows to get rid of both powers of x in a reasonable way.
I did not check if it works, however. The answer is full of powers of that, so it looks good.

 

[Svein]

try setting x=u⁶. The sqrt(x) = u³, cubed root(x) = u² and dx/du = 6u⁵. The rest is left to the student...