$\displaystyle \int \frac{\sqrt[3]{x} + \sqrt{x}}{\sqrt[6]{x} + 1} dx$
and also
$\displaystyle \int \frac{x+8}{(x-2)(x^2+1)} dx$
$\displaystyle \frac{\sqrt[3]x+\sqrt x}{\sqrt[6]x+1}$
$\displaystyle =\frac{\left(\sqrt[3]x+\sqrt x\right)\left(\sqrt[6]x-1\right)}{\left(\sqrt[6]x+1\right)\left(\sqrt[6]x-1\right)}$
$\displaystyle =\frac{x^{1/2}-x^{1/3}+x^{2/3}-x^{1/2}}{x^{1/3}-1}$
$\displaystyle =\frac{x^{2/3}-x^{1/3}}{x^{1/3}-1}$
$\displaystyle =\frac{x^{1/3}\left(x^{1/3}-1\right)}{x^{1/3}-1}$
$\displaystyle =x^{1/3}$
Now integrate.
Use partial fraction decomposition.$\displaystyle \int \frac{x+8}{(x-2)(x^2+1)} dx$
Edit: And make sure you are familiar with the integral $\displaystyle \int\!\frac{dx}{1+x^2}.$
very nice!
using your final answer as motivation, we could have just factored out $\displaystyle x^{1/3}$ from the numerator in the beginning. then everything else cancels in the next step. but seeing that that was a good move in the first place is probably hard, i guess.