Answer
\[x+3x^{\frac{2}{3}}+6x^{\frac{1}{3}}+6\ln\left|x^{\frac{1}{3}}-1\right|+C\]
Work Step by Step
Let \[I=\int\frac{\sqrt[3]{x}+1}{\sqrt[3]{x}-1}dx\]
\[I=\int\frac{x^{\frac{1}{3}}+1}{x^{\frac{1}{3}}-1}dx\;\;\;\ldots (1)\]
Substitute $\;x=t^3\;\;\;\ldots (2)$
\[\Rightarrow dx=3t^2 dt\]
(1) becomes
\[I=\int\left(\frac{t+1}{t-1}\right)3t^2dt\]
\[I=\int\left[\frac{(t-1)+2}{t-1}\right]3t^2dt\]
\[I=\int\left[1+\frac{2}{t-1}\right]3t^2dt\]
\[I=\int\left[3t^2+\frac{6t^2}{t-1}\right]dt\]
\[I=\int\left[3t^2+\frac{6(t^2-1+1)}{t-1}\right]dt\]
\[I=\int\left[3t^2+\frac{6(t-1)(t+1)+6}{t-1}\right]dt\]
\[I=\int\left[3t^2+6t+6+\frac{6}{t-1}\right]dt\]
\[I=t^3+3t^2+6t+6\ln|t-1|+C\]
Where $C$ is constant of integration
From (2)
\[I=x+3x^{\frac{2}{3}}+6x^{\frac{1}{3}}+6\ln\left|x^{\frac{1}{3}}-1\right|+C\]
Hence \[I=x+3x^{\frac{2}{3}}+6x^{\frac{1}{3}}+6\ln\left|x^{\frac{1}{3}}-1\right|+C\;\;\;.\]
