Answer
$x+3\sqrt[3]{x^{2}}+6\sqrt[3]{x}+6\ln|\sqrt[3]{x}-1|+C$
Work Step by Step
$I=\displaystyle \int\frac{\sqrt[3]{x}+1}{\sqrt[3]{x}-1}dx=\quad\left[\begin{array}{ll}
u=\sqrt[3]{x}, & u^{3}=x\\
& 3u^{2}du=dx
\end{array}\right]$
$=\displaystyle \int\frac{u+1}{u-1}3u^{2}du$
$=3\displaystyle \int\frac{u^{3}+u^{2}}{u-1}du$
Dividing $u^{3}+u^{2}$ with $u-1$ (synthetic division)
$\left[\begin{array}{lllll}
1| & 1 & 1 & 0 & 0\\
& & 1 & 2 & 2\\
& 1 & 2 & 2 & |2
\end{array}\right]$
$u^{3}+u^{2}=(u-1)(u^{2}+2u+2)+2,$
$\displaystyle \frac{u^{3}+u^{2}}{u-1}=u^{2}+2u+2+\frac{2}{u-1}$
$I=3\displaystyle \int(u^{2}+2u+2+\frac{2}{u-1})du$
$=u^{3}+3u^{2}+6u+6\ln|u-1|+C$
... bring back x
$=x+3\sqrt[3]{x^{2}}+6\sqrt[3]{x}+6\ln|\sqrt[3]{x}-1|+C$
