The 3 Integral $\int_0^\infty {x\,{\rm d}x\over \sqrt[3]{\,\left(e^{3x}-1\right)^2\,}}=\frac{\pi}{3\sqrt 3}\big(\log 3-\frac{\pi}{3\sqrt 3} \big)$ Hi I am trying evaluate this integral and obtain the closed form:$$
I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}  \right).
$$
The integral and result has all 3's everywhere.   I am not sure how to approach this on.  The denominator seems to be a problem.
If $\displaystyle x=\frac{2in\pi}{3}$ we have a singularity but I am not sure how to use complex methods. We will have  a branch cut because of the root function singularity.  
Differentiating under the integral sign did not help either.  I tried partial integration with $v=(e^{3x}-1)^{\frac{2}{3}}$ but this did not simplify since I get a power $x^n, \ (n>1)$ in the new integral.  
Thanks, how can we evaluate the integral I?  
 A: Substitute $e^{-x}=t$, then the integral can be written as:
\begin{align*}
  \int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &= -\int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt\tag 1
\end{align*}
Consider:
\begin{align*}
  I(a) &= \int_{0}^{1} \, \frac{t^{a+1}}{\left(1-t^3\right)^{2/3}}\, dt \\
  &= -\frac{1}{3} \, {\rm B}\left(\frac{1}{3}, \frac{a+2}{3}\right) \\
  I'(0) &= \int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt \\
  &= \frac{1}{9} \, {\left(\gamma + \psi\left(\frac{2}{3}\right)\right)} {\rm B}\left(\frac{1}{3}, \frac{2}{3}\right) \tag{2}
\end{align*}
Simplifying $(2)$ by using Gauss's digamma theorem for $m<k$
$\displaystyle \psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lfloor \frac{k-1}{2} \rfloor} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right)$
and Euler's reflection formula for gamma functions, 
$\displaystyle {\rm B}(1-z, z)= \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)} $
and from $(1)$,
\begin{align*}
\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &=  \frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}\right)
\end{align*}
A: Following @Pranav Arora's idea, if defining the following integral
$$ I(a)=\int_0^\infty\frac{\ln(1+at^3)}{1+t^3}dt, $$
the calculation will just be basic calculus without using advanced tools. In fact, $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^\infty\frac{t^3}{(1+t^3)(1+at^3)}dt\\
&=&\frac{1}{a-1}\int_0^\infty\left(\frac{1}{1+t^3}-\frac{1}{1+at^3}\right)dt\\
&=&\frac{1}{a-1}\left(\frac{2\pi}{3\sqrt3}-\frac{2\pi}{3\sqrt3\sqrt[3]a}\right)\\
&=&\frac{2\pi}{3\sqrt3}\frac{1}{\sqrt[3]a(1+\sqrt[3]a+\sqrt[3]a^2)}.
\end{eqnarray}
So
\begin{eqnarray}
I&=&\frac{2\pi}{3\sqrt3}\int_0^1\frac{1}{\sqrt[3]a(1+\sqrt[3]a+\sqrt[3]a^2)}da\\
&=&\frac{2\pi}{3\sqrt3}\int_0^1\frac{3u}{1+u+u^2}du\\
&=&\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}\right).
\end{eqnarray}
