$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$ Can anyone help me to cope with this integral? I have tried solving it but I had no breakthrough whatsoever ...
$$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$$

\begin{align}
\int_0^\infty x^5 e^{-x^3} \ln(1+x)\,dx 
&= -\frac{1}{3} e^{-x^3}\left(x^3+1\right)\ln(1+x)\Big|_0^\infty + \frac{1}{3} \int_0^\infty \frac{e^{-x^3}\left(x^3+1\right)}{1+x}\,dx \\
&= \frac{1}{3} \int_0^\infty \frac{e^{-x^3}\left(x^3+1\right)}{1+x}\,dx \\
&= \frac{1}{3} \int_0^\infty e^{-x^3}(x^2 - x + 1)\,dx \\
&= \frac{1}{3} \left[\int_0^\infty x^2e^{-x^3}\,dx - \int_0^\infty xe^{-x^3}\,dx + \int_0^\infty e^{-x^3}\,dx\right] \\
&= \frac{1}{3} \left[\frac{1}{3}\int_0^\infty e^{-u}\,du - \frac{1}{3}\int_0^\infty u^{-\frac{1}{3}}e^{-u}\,du + \int_0^\infty u^{-\frac{2}{3}}e^{-u}\,du\right] \\
&= \frac{1}{9} \left[1-\Gamma\left(\frac{2}{3}\right) + \Gamma\left(\frac{1}{3}\right)\right]
\end{align}
