Surprising Closed Form

Same solution as @Jack Lam , presented in more detail.

$\begin{aligned} I & = \int_0^\infty \frac {xe^\frac x{10}}{100(e^\frac x5 + 1)} dx & \small \color{#3D99F6}{\text{Let }u=e^\frac x{10} \implies x = 10 \ln u, \ dx = \frac {10}u du} \\ & = \int_1^\infty \frac {10 \ln u \cdot u}{100(u^2 + 1)} \cdot \frac {10}u du \\ & = \int_1^\infty \frac {\ln u}{u^2 + 1} du \\ & = \color{#3D99F6}{G} \approx \boxed{0.915966} \end{aligned}$

where $\color{#3D99F6}{G}$ is Catalan's constant .

Let

$I = \displaystyle \int_0^{\infty} \dfrac{x}{100} \cdot \color{#3D99F6}{\dfrac{e^{\frac{x}{10}}}{\left( e^{\frac x5} +1 \right)}} \, dx$

Lets manipulate the part highlighted in blue. Assume that $\dfrac{x}{10} = t$ , so that part becomes

$\begin{aligned} \dfrac{e^t}{e^{2t}+1} &= \dfrac{1}{e^t + \frac{1}{e^t}} \\ &= \dfrac{1}{e^t + e^{-t}} \\ &= \dfrac 12 \cdot \dfrac{2}{e^t + e^{-t}} \\ &= \dfrac 12 \cdot \dfrac{1}{\cosh(t)} \\ &= \dfrac 12 \cdot \dfrac{1}{\cosh \left( \frac{x}{10} \right)} \end{aligned}$

Thus, our integral is

$\begin{aligned} I &= \displaystyle \int_0^{\infty} \dfrac{x}{100} \cdot \dfrac 12 \cdot \dfrac{1}{\cosh \left( \frac{x}{10} \right)} \, dx \\ &= \dfrac{1}{10} \left[ \dfrac 12 \displaystyle \int_0^{\infty} \dfrac{\left( \frac{x}{10} \right)}{\cosh \left( \frac{x}{10} \right)} \, dx \right] \\ &= \dfrac{1}{10} \cdot 10G \qquad \color{#D61F06}{(\star)} \\ &= G \\ &= \boxed{0.91} \end{aligned}$

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One of the many known integral representations of $G$ ( Catalan's Constant ) include

$G = {\dfrac {1}{2}}\int _{0}^{\infty }{\frac {t}{\cosh t}}\,dt$

The substitution $u = e^{\frac{x}{10}}$ or $u = e^{\frac{-x}{10}}$ transforms it into one of the many integral definitions of Catalan's Constant .