How do you evaluate $\int _1^e\:\frac{\ln(x)}{x+1}\,dx$

Question

How do you evaluate
$$ \large\int\limits _1^e\frac{\ln(x)}{x+1}\;\mathrm{d}x $$

Answer 1

$$ \begin{align} \int_1^e\frac{\log(x)}{x+1}\,\mathrm{d}x &=\int_0^1\frac{x}{1+e^{-x}}\,\mathrm{d}x\\ &=\int_0^1x\left(\sum_{k=0}^\infty(-1)^ke^{-kx}\right)\,\mathrm{d}x\\ &=\frac12+\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\int_0^kxe^{-x}\,\mathrm{d}x\\ &=\frac12+\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\left(1-(k+1)e^{-k}\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac12-\frac{\pi^2}{12}+\log\left(1+e^{-1}\right)-\mathrm{Li}_2\left(-e^{-1}\right)}\tag{1} \end{align} $$ 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.

Answer 2

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

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

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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!