Answer
$3\displaystyle \ln|x^{1/3}-1|+\frac{3x^{2/3}}{2}+3x^{1/3}+x+C_{1}$
Work Step by Step
sub: $\left[\begin{array}{ll}
u=x^{1/3}-1 & \\
du=\frac{1}{3x^{2/3}}dx & dx=3(u+1)^{2}du
\end{array}\right]$
$\displaystyle \int\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}dx=\int\frac{u+1}{u}3(u+1)^{2}du$
$=3\displaystyle \int\frac{u+1}{u}(u^{2}+2u+1)du$
$=3\displaystyle \int(u^{2}+3u+3+\frac{1}{u})du$
$=3[\displaystyle \frac{u^{3}}{3}+\frac{3u^{2}}{2}+3u+\ln|u|]+C$
$=3[\displaystyle \frac{(x^{1/3}-1)^{3}}{3}+\frac{3(x^{1/3}-1)^{2}}{2}+3(x^{1/3}-1)+ \ln|x^{1/3}-1|]+C\\$
$=x^{3/3}-3x^{2/3}+3x^{1/3}+\displaystyle \frac{9}{2}x^{2/3}-9x^{1/3}+\frac{9}{2}+9x^{1/3}-9 \\+3\ln|x^{1/3}-1|+C$
$=3\displaystyle \ln|x^{1/3}-1|+\frac{3x^{2/3}}{2}+3x^{1/3}+x+C_{1}$
