Answer
$$\int\frac{1}{(x^{1/3}-1)\sqrt x}dx=6x^{1/6}+3(\ln|x^{1/6}-1|-\ln|x^{1/6}+1|)+C$$
Work Step by Step
$$I=\int\frac{1}{(x^{1/3}-1)\sqrt x}dx$$
We set $u^6=x$, so $$6u^5du=dx$$
Also, $\sqrt x=u^3$ and $x^{1/3}=u^2$
$$I=\int\frac{6u^5}{(u^2-1)u^3}du$$ $$I=\int\frac{6u^2}{u^2-1}du=\int\frac{6u^2-6}{u^2-1}du+\int\frac{6}{u^2-1}du$$ $$I=\int6du+3\int\frac{2}{(u-1)(u+1)}du$$ $$I=6u+3\int\frac{(u+1)-(u-1)}{(u-1)(u+1)}du$$ $$I=6u+3\Big(\int\frac{du}{u-1}-\int\frac{du}{u+1}\Big)$$ $$I=6u+3(\ln|u-1|-\ln|u+1|)+C$$ $$I=6x^{1/6}+3(\ln|x^{1/6}-1|-\ln|x^{1/6}+1|)+C$$
