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NonCommAlg first two simple facts that we will use them later on in the solution:
$\displaystyle (*)$ if $\displaystyle y > 0$, then $\displaystyle \int_0^{\infty}\frac{dt}{t^3+y}=\frac{1}{\sqrt[3]{y^2}} \int_0^{\infty}\frac{dt}{1+t^3},$ and $\displaystyle \int_0^{\infty} \frac{dt}{1+t^3} = \frac{2\pi}{3\sqrt{3}}.$
now, in your integral, put $\displaystyle e^{3x}-1=\frac{1}{t^3}.$ then $\displaystyle x=\frac{1}{3}\ln\left(1+\frac{1}{t^3}\right),$ and $\displaystyle dx=\frac{-dt}{t(1+t^3)}.$ thus:
$\displaystyle I=\frac{1}{3}\int_0^{\infty}\frac{\ln(1+\frac{1}{t ^3})}{1+t^3} \ dt=\frac{1}{3}\int_0^{\infty} \int_0^1\frac{1}{(t^3+y)(t^3+1)} \ dy \ dt $
$\displaystyle =\frac{1}{3} \int_0^1 \int_0^{\infty} \frac{1}{(t^3+y)(t^3+1)} \ dt \ dy=\frac{1}{3}\int_0^1 \frac{1}{1-y}\int_0^{\infty} \left(\frac{1}{t^3+y}-\frac{1}{t^3+1}\right) \ dt \ dy.$ thus by $\displaystyle (*)$:
$\displaystyle I=\frac{2\pi}{9\sqrt{3}}\int_0^1 \frac{1-\sqrt[3]{y^2}}{\sqrt[3]{y^2}(1-y)} \ dy.$ now let $\displaystyle y=z^3.$ then we will have:
$\displaystyle I=\frac{2\pi}{3\sqrt{3}} \int_0^1 \frac{1+z}{1+z+z^2} \ dz=\frac{2\pi}{3\sqrt{3}}\left[\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2z+1}{\sqrt{3}}\right)+\frac{1}{2}\l n(1+z+z^2) \right]_ 0^1$
$\displaystyle =\frac{\pi}{3\sqrt{3}}\left(\ln 3 + \frac{\pi}{3\sqrt{3}} \right). \ \ \ \ \square $