<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] Take 
$$t = (e^x - 1)^{1/3}$$
or
$$x = \ln( t^3 + 1 )$$
Then 
$$dt = \frac{1}{3}e^x(e^x-1)^{-2/3} \ dx$$
Hence 
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}
=
3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{1 }{t e^x(e^x-1)^{-2/3}}\ dt
=
3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{t}{(t^3 + 1)}\ dt
$$
The last integral could be written as
$$
{\displaystyle\int} _{0}^{(e-1)^{1/3}}\dfrac{t}{\left(t+1\right)\left(t^2-t+1\right)}\,\mathrm{d}t$$
Perform Partial Fraction Decomposition
$${\displaystyle\int} _{0}^{(e-1)^{1/3}}\left(\dfrac{t+1}{3\left(t^2-t+1\right)}-\dfrac{1}{3\left(t+1\right)}\right)\mathrm{d}t$$
Take the first integral above, i.e. ${\displaystyle\int}\dfrac{t+1}{t^2-t+1}\,\mathrm{d}t$ could be written as 
$$\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{2t-1}{t^2-t+1}\,\mathrm{d}t+\class{steps-node}{\cssId{steps-node-8}{\dfrac{3}{2}}}{\displaystyle\int}\dfrac{1}{t^2-t+1}\,\mathrm{d}t$$
which is equal to (with some straight forward math)
$$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$$
Now, we need $\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$
which is easily shown to be 
$$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)==\ln\left(t+1\right)$$
Finalizing we get
$$3\left(-\dfrac{\ln\left(\left|t+1\right|\right)}{3}+\dfrac{\ln\left(t^2-t+1\right)}{6}+\dfrac{\arctan\left(\frac{2t-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)\Big\vert_0^{(e-1)^{1/3}}$$
which is
$$3\left(\dfrac{\ln\left(\left(\mathrm{e}-1\right)^\frac{2}{3}-\sqrt[3]{\mathrm{e}-1}+1\right)}{6}+\dfrac{\arctan\left(\frac{2\cdot\sqrt{3}\sqrt[3]{\mathrm{e}-1}-\sqrt{3}}{3}\right)}{\sqrt{3}}-\dfrac{\ln\left(\sqrt[3]{\mathrm{e}-1}+1\right)}{3}+\dfrac{{\pi}}{2{\cdot}3^\frac{3}{2}}\right)$$
</s>