Problem finding $\int\sqrt[3]{\frac{x+1}{x-1}}dx$ Given the integral  $\int\sqrt[3]{\frac{x+1}{x-1}}dx$, I can't find its value. I can't one correct substitution. I tried $t=\sqrt[3]{x+1}$ but then $dt\cdot {3\sqrt[3]{x+1}^2}=dx$ but then I'm stuck with integrating $\frac{3t^3}{\sqrt[3]{t^3-2}}$.Almost the same when substituing $t=\sqrt[3]{x-1}$. I also tried multiplying and diving by $\sqrt[3]{(\frac {x+1}{x-1})^{2/3}}$ but again 3rd root ruins everything. Which substitution will solve the integral?

Hint: Put $$\frac{x+1}{x-1}=u^3$$ Then, $$dx=\frac{3u^2(u^3-1)-3u^2(u^3+1)}{(u^3-1)^2}du=\frac{-6u^2}{(u^3-1)^2}du$$the integral becomes $$-\int \frac{6u^3}{(u^3-1)^2}du $$Then use partial fraction expansion.
