Showing $\int_{0}^{\infty}\frac{[\ln(x+1)]^{4}}{x^{2}}dx=\frac{4\pi^{4}}{15}$

Question

How to solve this weird integral?

$$\int_{0}^{\infty}\frac{[\ln(x+1)]^{4}}{x^{2}}dx$$

I know the answer is $\dfrac{4\pi^{4}}{15}$, but I don't know how to get it. I tried to use Gamma-function with no success.

Answer 1

Performing the substitution $t=\frac{1}{1+x}$ yields $$\int _0^{\infty }\frac{\ln ^4\left(1+x\right)}{x^2}\:\mathrm{d}x=\int _0^1\frac{\ln ^4\left(t\right)}{\left(1-t\right)^2}\:\mathrm{d}t$$ Now expand the denominator $$=\sum _{n=1}^{\infty }n\int _0^1t^{n-1}\ln ^4\left(t\right)\:\mathrm{d}t=24\sum _{n=1}^{\infty }\frac{1}{n^4}$$ $$\therefore \int _0^{\infty }\frac{\ln ^4\left(1+x\right)}{x^2}\:\mathrm{d}x=\frac{4\pi ^4}{15}$$