How do you evaluate $\int _1^e\:\frac{\ln(x)}{x+1}\,dx$ How do you evaluate
$$
  \large\int\limits _1^e\frac{\ln(x)}{x+1}\;\mathrm{d}x
$$
 A: $$
\begin{align}
\int_1^e\frac{\log(x)}{x+1}\,\mathrm{d}x
&=\int_0^1\frac{x}{1+e^{-x}}\,\mathrm{d}x\\
&=\int_0^1x\left(\sum_{k=0}^\infty(-1)^ke^{-kx}\right)\,\mathrm{d}x\\
&=\frac12+\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\int_0^kxe^{-x}\,\mathrm{d}x\\
&=\frac12+\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\left(1-(k+1)e^{-k}\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12-\frac{\pi^2}{12}+\log\left(1+e^{-1}\right)-\mathrm{Li}_2\left(-e^{-1}\right)}\tag{1}
\end{align}
$$
Formula $(5)$ from this answer proves the Inversion Formula for $\mathrm{Li}_2$:
$$
\mathrm{Li}_2(x)+\mathrm{Li}_2(1/x)=-\frac{\pi^2}6-\frac12\log(-x)^2\tag{2}
$$
Formula $(2)$, with $x=-e$, allows $(1)$ to be rewritten as
$$
\int_1^e\frac{\log(x)}{x+1}\,\mathrm{d}x
=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^2}{12}+\log(e+1)+\mathrm{Li}_2(-e)}\tag{3}
$$
which is the answer Dr. MV got.
A: Integration by parts gives
$$\begin{align}
\int_1^e \frac{\ln x}{x+1}dx&=\left.\left(\ln x\ln(1+x)\right)\right|_{1}^{e}-\int_0^e\frac{\ln (1+x)}{x}dx\\\\
&=\ln(1+e) -\int_1^e\frac{\ln (1+x)}{x}dx\tag 1
\end{align}$$
Next, we recall that the Dilogarithm function $\text{Li}_2$ can be written as
$$\begin{align}
\text{Li}_2(x)&=-\int_0^x\frac{\ln(1-u)}{u}du\\\\
&=-\int_0^{-x}\frac{\ln(1+u)}{u}du\tag 2
\end{align}$$
whereupon substituting $(2)$ into $(1)$ yields
$$\int_1^e \frac{\ln x}{x+1}dx=\ln(1+e)+\text{Li}_2(-e)-\text{Li}_2(-1) \tag 3$$
Note that $\text{Li}_2(-1)=-\eta(2)=-\frac12 \zeta(2)=-\frac{\pi^2}{12}$, 
where $\eta(x)$ is the Dirichlet Eta function and $\zeta(x)$ is the Riemann Zeta function.  Putting it all together reveals that 

$$\int_1^e \frac{\ln x}{x+1}dx=\ln(1+e)+\text{Li}_2(-e)+\frac{\pi^2}{12} \tag4$$


NOTE:
We can use the inversion formula for  the Dilogarithm function 
$$\text{Li}_2(z)=-\frac12(\ln(-z))^2-\pi^2/6-\text{Li}_2(1/z)$$
to recast $(4)$ as

$$\int_1^e \frac{\ln x}{x+1}dx=\frac12+\ln(1+e^{-1})-\frac{\pi^2}{12}-\text{Li}_2(-e^{-1})$$

which agrees with the result reported by @robjon!
