<s>[INST] Evaluate $\int _0^1\frac{dx}{(e^x-1)^{1/3}}$ Study the convergence of the following integral for $\alpha\in\mathbb{R}$ and evaluate it for $\alpha = \frac{1}{3}$ if it converges for that value.
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\alpha}} x$$
The integral converges for $\alpha \lt 1$, if I am not wrong, so I can evaluate:
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}$$
whose primitive, using some razionalizations and change of variables, is: $\frac{1}{2}\log|4(e^x-1)^\frac{2}{3}-4(e^x-1)^\frac{1}{3}+4|+\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\left(2(e^x-1)^\frac{1}{3}-1\right)\right)-\log|(e^x-1)^\frac{1}{3}+1|+C$
Anyone knows if there is a less intricate way in order of obtain it than almost 3 sheet full of calculations? Thank you
 [/INST] using $u={{\left( {{e}^{x}}-1 \right)}^{1/3}}$ leads to $\int{\frac{3u}{{{u}^{3}}+1}du}$
$$
\begin{align}
  & =\int{-\frac{1}{u+1}+\frac{u+1}{{{u}^{2}}-u+1}du} \\ 
 & =\int{-\frac{1}{u+1}du}+\frac{1}{2}\int{\frac{2u-1}{{{u}^{2}}-u+1}du}+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\ 
 & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\ 
\end{align}
$$
For the last integral use $v=2u-1$ and notice that $4\left( {{u}^{2}}-u+1 \right)={{v}^{2}}+3$
$$
\begin{align}
  & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\int{\frac{1}{{{v}^{2}}+3}dv} \\ 
 & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\frac{1}{\sqrt{3}}ta{{n}^{-1}}\left( \frac{2u-1}{\sqrt{3}} \right) \\ 
\end{align}
$$
</s>