$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$

Question

Can anyone help me to cope with this integral? I have tried solving it but I had no breakthrough whatsoever ...

$$\int_0^\infty x^5e^{-x^3}\ln(1+x)dx$$

Answer 1

\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}