Nice solution mathofusva! Here is my attempt for $n>2$.
\begin{align*}
\int_{1}^{+\infty}{\frac{\log(x+1)}{x^n}\,dx} &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}}\\
{dx\,=\,-\frac{1}{t^2}dt}\\
\end{subarray}}\,-\int_{1}^{0}{t^n\log\big(\tfrac{1}{t}+1\big)\frac{1}{t^2}\,dt}\\
&=\int_{0}^{1}{t^{n-2}\log\big(\tfrac{1}{t}+1\big)\,dt}\\
&=\int_{0}^{1}{\Big(\frac{t^{n-1}}{n-1}\Big)'\log\big(\tfrac{1}{t}+1\big)\,dt}\\
&=\Big[\frac{t^{n-1}}{n-1}\log\big(\tfrac{1}{t}+1\big)\Big]_{0}^{1}+\frac{1}{n-1}\int_{0}^{1}{\frac{t^{n-2}}{t+1}\,dt}\\
&=\frac{\log2}{n-1}-\frac{1}{n-1}\,\cancelto{0}{\mathop{\lim}\limits_{t\to0^{+}}t^{n-1}\log\big(\tfrac{1}{t}+1\big)}\;+\frac{1}{n-1}\int_{0}^{1}{t^{n-2}\mathop{\sum}\limits_{m=0}^{+\infty}(-1)^mt^m\,dt}\\
&=\frac{\log2}{n-1}+\frac{1}{n-1}\mathop{\sum}\limits_{m=0}^{+\infty}(-1)^m\int_{0}^{1}{t^{m+n-2}\,dt}\\
&=\frac{\log2}{n-1}+\frac{1}{n-1}\mathop{\sum}\limits_{m=0}^{+\infty}\frac{(-1)^m}{m+n-1}\\
&=\frac{\log2}{n-1}+\frac{1}{2\,(n-1)}\Big(\psi\big(\tfrac{n}{2}\big)-\psi\big(\tfrac{n-1}{2}\big)\Big)\,,
\end{align*}
from which
this question came up.