Find the value of $\int_{1}^{e} \frac{\ln x}{x+1}dx$ 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?
 A: $$\begin{eqnarray*}\int_{0}^{1}\frac{t}{1+e^{-t}}\,dt &=& \frac{1}{2}+\sum_{n\geq 1}(-1)^n\int_{0}^{1}t e^{-nt}\,dt\\[0.2cm]&=&\frac{1}{2}+\sum_{n\geq 1}(-1)^n\frac{1-(n+1)e^{-n}}{n^2}\\[0.2cm]&=&\frac{1}{2}-\frac{\pi^2}{12}+\log\left(1+\frac{1}{e}\right)-\text{Li}_2\left(-\frac{1}{e}\right)\\[0.2cm]&=&\color{red}{\frac{\pi^2}{12}+\log(1+e)+\text{Li}_2(-e)}\\[0.2cm]&=&0.32944265\ldots \end{eqnarray*}$$
A: Make substitution $u= ln(x)$, $e^u = x \rightarrow e^u du =
dx$ . We then must evaluate 
$$ \int_{1}^{e} \frac{ue^u}{e^u+1} $$
We invoke Integration by parts,
$$ \int fg' = fg - \int f'g$$
where $f = u, g'= \frac{e^u}{e^u+1}$
Giving us
$$ u \int \frac{e^u}{e^u+1} du- \int \int  \frac{e^u}{e^u+1} du \ du $$
To evaluate
$$ \int \frac{e^u}{e^u+1}  du $$
Let $r = e^u + 1, dr = e^u$ gives us
$$ \int \frac{1}{r} dr $$
Which is $ \ln(r) = \ln(e^u+1)$
Thus:
$$ u \int \frac{e^u}{e^u+1} du- \int \int  \frac{e^u}{e^u+1} du \ du $$
Reduces to:
$$ u \ln(e^u+1) - \int  \ln(e^u+1) du $$
This isn't immediately obvious (hit with Wolfram) to integrate but yields:
$$ u \ln(e^u+1) + Li_2(e^{-u}) $$
Which itself reduces to
$$ \ln(x)\ln(x+1) + Li_2(\frac{1}{x})$$
We can then take difference evaluation at e and then 1 and subtracting.
A: \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}
