Evaluate: $$\int_{0}^{1} \frac{\ln^k x}{x+1}$$
It looks like the only way to start this is to use that
$$\frac{1}{x+1}= \sum_{n = 0}^{\infty} (-1)^n x^n $$
Then we have to evaluate
$$\int_{0}^{1} {\ln^k x}\sum_{n = 0}^{\infty} (-1)^n x^n = \int_{0}^{1} \sum_{n = 0}^{\infty} (-1)^n x^n{\ln^k x}=\sum_{n = 0}^{\infty} \int_{0}^{1}(-1)^n x^n \ln^kx$$
For $k = 1$, it comes to be $\frac{\pi^2}{12}$
For $k = 3$, it comes to be $\frac{7\pi^4}{720}$
So, there is $\zeta(k+1)$ involved.
EDIT: After trying for different values of $k$ it seems like the integral evaluates to be $(1 - \frac{1}{2^k}) \zeta(k+1)$
Any help would be appreciated.
