Show that $\int_{1/e}^e \left|\frac{\ln x}{1+x}\right|\,\mathrm{d}x=\frac{1}{2}$ How can one show that

$$I=\int_{1/e}^e \left|\frac{\ln x}{1+x}\right|\,\mathrm{d}x=\frac{1}{2}$$

preferably without resorting to the polylogarithm function?
Mathematica returns
1 + Pi^2/6 + PolyLog[2, -1/E] + PolyLog[2, -E]

which is numerically equivalent to $0.5$.
The first thing that comes to mind is to remove the absolute value and consider the two integrals

$$I=\color{red}{I_1}-\color{blue}{I_2}=\color{red}{\int_1^e\frac{\ln x}{1+x}\,\mathrm{d}x}-\color{blue}{\int_{1/e}^1\frac{\ln x}{1+x}\,\mathrm{d}x}$$

Is there a substitution that would reduce this to $\frac{1}{2}$?
Another method I've considered is to integrate by parts:
$$I=1-\int_1^e\frac{\ln(1+x)}{x}\,\mathrm{d}x+\int_{1/e}^1\frac{\ln(1+x)}{x}\,\mathrm{d}x$$
but this only seems to force me down a path of using the Mercator series (at least for the second integral, considering the interval of convergence of the series) and polylogarithm.
 A: By setting $x=e^t$ we are left with
$$ I = \int_{-1}^{1}e^t\left|\frac{t}{1+e^t}\right|\,dt = \int_{0}^{1}\frac{t}{1+e^{-t}}\,dt+\int_{0}^{1}\frac{t}{1+e^t}\,dt=\int_{0}^{1}t\,dt=\color{red}{\frac{1}{2}},$$
sic et simpliciter.
A: Jack D'Aurizio posted a simpler answer above, but since I was already in the process of typing: 
Make the substitution $x \mapsto 1/x$ to see $$\int_{1/e}^1 \frac{\log x}{1+x} dx= \int^1_{e} \frac{\log(1/x)}{1+1/x} \left( -\frac 1 {x^2} \right)dx = -\int^e_1 \frac{\log(x)}{x^2+x}dx.$$ Thus $$I = \int^e_1 \frac{\log x}{x+1} + \frac{\log x}{x^2+x} dx = \int^e_1 \frac{(x+1)\log(x)}{x^2+x}dx= \int^e_1 \frac{\log x}{x}dx = \left.\frac{1}{2}\log(x)^2\right|^e_1 = \frac 1 2.$$ 
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I & =
\color{#f00}{\int_{1/\expo{}}^{\expo{}}\verts{\ln\pars{x} \over 1 + x}\,\dd x}
\,\,\,\stackrel{x\ \equiv\ \expo{-t}}{=}\,\,\,
\int_{1}^{-1}\verts{-t \over \expo{-t} + 1}\pars{-\expo{-t}}\,\dd x =
\int_{-1}^{1}{\verts{t} \over \expo{t} + 1}\,\dd x
\\[5mm] & =
\int_{-1}^{1}\verts{t}\bracks{\Theta\pars{-t} + {\mrm{sgn}\pars{t} \over \expo{\verts{t}} + 1}}\,\dd x\qquad\qquad
\pars{
\begin{array}{rll}
\ds{\Theta} & \ds{:} & Heaviside\ Step\ Function
\\
\ds{\mrm{sgn}} & \ds{:} & Sign\ Function
\end{array}}
\\[5mm] & =
\int_{-1}^{0}\pars{-t}\,\dd t = \color{#f00}{1 \over 2}
\end{align}

Note that
  $\ds{\,\mrm{f}\pars{x} \equiv {1 \over \expo{x} + 1} =
1 - {1 \over \expo{-x} + 1} = 1 - \mrm{f}\pars{-x}}$ which satisfies:
  \begin{align}
\mrm{f}\pars{x} & =
\left\{\begin{array}{lcrcl}
\ds{1 - {1 \over \expo{\verts{x}} + 1}} & \mbox{if} & \ds{x} & \ds{<} & \ds{0}
\\
\ds{1 \over 2} & \mbox{if} & \ds{x} & \ds{=} & \ds{0}
\\
\ds{1 \over \expo{\verts{x}} + 1} & \mbox{if} & \ds{x} & \ds{>} & \ds{0}
\end{array}\right.
\\[5mm]
\mbox{It's equivalent to}\
\mrm{f}\pars{x} & =
\left\{\begin{array}{lcrcl}
\ds{\Theta\pars{-x} + {\mrm{sgn}\pars{x} \over \expo{\verts{x}} + 1}} & \mbox{if} & \ds{x} & \ds{\not=} & \ds{0}
\\
\ds{1 \over 2} & \mbox{if} & \ds{x} & \ds{=} & \ds{0}
\end{array}\right.
\end{align}

A: $$
\begin{align}
\int_{1/e}^e\left|\frac{\log(x)}{1+x}\right|\,\mathrm{d}x
&=\int_1^e\frac{\log(x)}{1+x}\,\mathrm{d}x-\int_{1/e}^1\frac{\log(x)}{1+x}\,\mathrm{d}x\\
&=\int_1^e\frac{\log(x)}{1+x}\,\mathrm{d}x+\int_1^e\frac{\log(x)}{x(1+x)}\,\mathrm{d}x\\
&=\int_1^e\frac{\log(x)}{x}\,\mathrm{d}x\\
&=\left[\frac12\log(x)^2\right]_1^e\\[3pt]
&=\frac12
\end{align}
$$
