TITLE: Log integrals I
QUESTION [13 upvotes]: In this example the value of the integral 
\begin{align}
I_{3} = \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx 
\end{align}
was derived. The purpose of this question is to determine the value of the more general form of the integral, namely,
\begin{align}
I_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{x} \, dx . 
\end{align}
Also is there a corresponding value for the integral
\begin{align}
J_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{1+x} \, dx \hspace{3mm} ?
\end{align}

REPLY [4 votes]: As M.N.C.E. has provided a nice answer for $I_{n}$ the only change that could be made it to cast the result into the form
\begin{align}
I_{n} &= \int_{0}^{1} \frac{\ln^{n}(1+x)}{x} \, dx = - \frac{n}{n+1} \, \ln^{n+1}(2) + n! \zeta(n+1) - \sum_{k=1}^{n} \frac{n! \, \ln^{n-k}(2)}{(n-k)!} \, Li_{k+1}\left(\frac{1}{2}\right).
\end{align}
The second integral is
\begin{align}
J_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{1+x} \, dx = \left[ \frac{\ln^{n+1}(1+x)}{n+1} \right]_{0}^{1} = \frac{\ln^{n+1}(2)}{n+1}
\end{align}