Evaluating $\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx $ I wonder if this integral $$\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx \quad n=1,2,3,...$$ admits a general formula for integers $n$. I've found $$\int_{0}^{1}\frac{\ln{(1+x)}}{1+x} dx = \left[ \frac{1}{2}\left(\ln(1+x)\right)^2 \right]_0^1=\frac{1}{2}\left(\ln2\right)^2. $$ Using integration by parts twice,
$$\begin{align}
\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^2 dx &= \left[ \frac{-1}{(1+x)}\left(\ln(1+x)\right)^2 \right]_0^1-\int_{0}^{1}\frac{\ln{(1+x)}}{\left(1+x\right)^2} dx \\
&=-\frac{1}{2}\left(\ln2\right)^2+ \left[ \frac{1}{(1+x)}\ln(1+x) \right]_0^1-\int_{0}^{1}\frac{1}{\left(1+x\right)^2} dx\\
&=-\frac{1}{2}\left(\ln2\right)^2+\frac{1}{2}\ln2-\frac{1}{2}.
\end{align}$$
Could you find a general formula? Any help is welcome! Thank you.
 A: Result:
$$\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x=\frac{n!}{(n-1)^{n+1}}-\frac{n!}{2^{n-1}}\sum^n_{j=0}\frac{\ln^{j}{2}}{j!(n-1)‌​^{n-j+1}}$$
$\text{for $n\in \mathbb{N}$, $n\geq2$}.$

Derivation:
\begin{align}
\small{\int^1_0\frac{\ln^n(1+x)}{(1+x)^n}{\rm d}x}
&\small{=\int^2_1\frac{\ln^n{x}}{x^n}{\rm d}x\tag1}\\
&\small{=(-1)^n\int^1_{\frac{1}{2}}x^{n-2}\ln^n{x}\ {\rm d}x\tag2}\\
&\small{=(-1)^n\frac{\partial^n}{\partial n^n}\int^1_{\frac{1}{2}}x^{n-2}{\rm d}x}\\
&\small{=(-1)^n\frac{\partial^n}{\partial n^n}\left[\frac{1}{n-1}-\frac{1}{2^{n-1}(n-1)}\right]}\\
&\small{=(-1)^n\left[\frac{(-1)^nn!}{(n-1)^{n+1}}-\sum^n_{j=0}\binom{n}{j}\frac{(-1)^{n-j}(n-j)!}{(n-1)^{n-j+1}}\frac{(-1)^{j}\ln^{j}(2)}{2^{n-1}}\right]\tag3}\\
&\small{=\frac{n!}{(n-1)^{n+1}}-\frac{n!}{2^{n-1}}\sum^n_{j=0}\frac{\ln^{j}{2}}{j!(n-1)‌​^{n-j+1}}}
\end{align}

Explanation:
$(1):\text{Let $x\mapsto x-1$}$
$(2):\text{Let $x\mapsto x^{-1}$}$
$(3):\text{Apply Leibniz Generalised Product Rule}$
A: \begin{align}\int^{1}_{0}\Big(\frac{\ln(1+x)}{1+x}\Big)^n\,dx&=\int^{\ln(2)}_{0}\Big(\frac{u}{e^u}\Big)^ne^u\,du\\&=\int^{\ln(2)}_{0}u^ne^{-(n-1)u}\,du=\frac{1}{(n-1)^{n+1}}\int^{(n-1)\ln(2)}_{0}v^ne^{-v}\,dv\\&=\frac{1}{(n-1)^{n+1}}\Big(\int^{\infty}_{0}v^ne^{-v}\,dv-\int^{\infty}_{(n-1)\ln(2)}v^ne^{-v}\,dv\Big)\\&=\frac{1}{(n-1)^{n+1}}\Big(\Gamma(n+1)-\Gamma(n+1,(n-1)\ln(2))\Big)\\&=\frac{n!}{(n-1)^{n+1}}-\frac{\Gamma(n+1,(n-1)\ln(2))}{(n-1)^{n+1}}\end{align}
where $\Gamma(n,x)$ is the upper incomplete gamma function.
