TITLE: Find the value of $\int_{1}^{e} \frac{\ln x}{x+1}dx$
QUESTION [5 upvotes]: Find the value of
$$\int_{1}^{e} \frac{\ln x}{x+1}dx.$$
I don't have solution for this problem. Can you help me?

REPLY [2 votes]: \begin{align}
\int_{1}^{e} \frac{\ln x}{x+1}dx&=\int_1^e\int_{y=1}^x\frac{1}{y}\frac{1}{x+1}dydx\\
&=\int_1^e\int_{x=y}^e\frac{1}{y}\frac{1}{x+1}dxdy\\
&=\int_1^e\frac{\log(e+1)-\log(y+1)}{y}dy\\
&=\int_1^e\frac{\log(e+1)}{y}dy-\int_1^e\frac{\log(y+1)}{y}dy\\
&=\log(e+1)-\int_1^e\frac{\log(y+1)}{y}dy\\
&=\log(e+1)-\int_1^e\sum_{j=0}^{\infty}\frac{(-y)^j}{j+1}dy\\
&=\log(e+1)-\sum_{j=0}^{\infty}(-1)^j\frac{e^{j+1}-1}{(j+1)^2}\\
&=\log(e+1)+\sum_{j=0}^{\infty}\frac{(-e)^{j+1}}{(j+1)^2}-\sum_{j=0}^{\infty}(-1)^{j+1}\frac{1}{(j+1)^2}\\
&=\log(e+1)+\sum_{j=1}^{\infty}\frac{(-e)^j}{j^2}-\sum_{j=1}^{\infty}(-1)^j\frac{1}{j^2}\\
&=\color{blue}{\log(e+1)+\text{Li}_2(-e)+\frac{\pi^2}{12}}
\end{align}