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?  

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}
