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We wish to prove that

$$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}\tag1$$

$$1-e^{3x}=(1-e^x)(1+e^x+e^{2x})\tag2$$

$$1+e^{3x}=(1+e^x)(1-e^x+e^{2x})\tag3$$

Sub $(2)$ and $(3)$ into $(1)\rightarrow (4)$

$$I=\int_{0}^{\infty}\left(2\cdot{1\over 1+e^x+e^{2x}}+{1\over 1-e^x+e^{2x}}\right)dx\tag5$$

Any hint, please, I am unable to continue.

$$I=\int_{0}^{1}\left({2u\over u^2+u+1}+{u\over u^2-u+1}\right)du=I_1+I_2\tag6$$

Respectively.

Apply formula (17) to $(6)$

Hence

$$I_1=\ln{3}-{2\sqrt3\over 3}\tan^{-1}\left({3\over \sqrt3}\right)$$

$$I_2={2\over \sqrt3}\tan^{-1}\left({\sqrt3 \over 3}\right)$$

Hence $I=\ln{3}$

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Any hint, please, I am unable to continue.

Hint. By the change of variable $u=e^{-x}$, $dx=-\dfrac{du}u$, one gets $$ \int_{0}^{\infty}\left(2\cdot{1\over 1+e^x+e^{2x}}+{1\over 1-e^x+e^{2x}}\right)dx=\int_{0}^1\left(2\cdot{u\over u^2+u+1}+{u\over u^2-u+1}\right)du $$ which is standard to evaluate.

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{I} & = \int_{0}^{\infty}\pars{2\,{1 - \expo{x} \over 1 - \expo{3x}} + {1 + \expo{x} \over 1 + \expo{3x}}}\,\dd x\ \stackrel{x\ =\ -\ln\pars{t}}{=}\ \int_{1}^{0}\pars{2\,{1 - 1/t \over 1 - 1/t^{3}} + {1 + 1/t \over 1 + 1/t^{3}}} \,{\dd t \over -t} \\[4mm] & = \int_{0}^{1}\pars{{2t - 2t^{2} \over 1 - t^{3}} + {t + t^{2} \over 1 + t^{3}}}\,\dd t = \int_{0}^{1}{3t - t^{2} + t^{4} - 3 t^{5} \over 1 - t^{6}}\,\dd t \\[4mm] & \stackrel{t^{6}\ \mapsto\ t}{=}\ {1 \over 6}\int_{0}^{1}{3t^{-2/3}\ -\ t^{-1/2} + t^{-1/6} - 3 \over 1 - t} \,\dd t \\[4mm] & = -\,\half\int_{0}^{1}{1 - t^{-2/3} \over 1 - t}\,\dd t + {1 \over 6}\int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t - {1 \over 6}\int_{0}^{1}{1 - t^{-1/6} \over 1 - t}\,\dd t \\[4mm] & = -\,\half\bracks{\Psi\pars{1 \over 3} + \gamma} + {1 \over 6}\bracks{\Psi\pars{\half} + \gamma} - {1 \over 6}\bracks{\Psi\pars{5 \over 6} + \gamma} \end{align} $\ds{\gamma}$ is the Euler-Mascheroni Constant.


Digamma Function $\ds{\Psi}$ with fractional arguments $\ds{0 < {p \over q} < 1}$ $\ds{\pars{~p = 1,2,3,\ldots\,,\quad p < q = 3,4,5,\ldots~}}$ are evaluated with the identity: $$ \Psi\pars{p \over q} + \gamma = H_{p/q} = -\ln\pars{2q} - {\pi \over 2}\,\cot\pars{p\pi \over q} + 2\sum_{k = 1}^{\left\lfloor\pars{q + 1}/2\right\rfloor - 1} \cos\pars{2kp\pi \over q}\ln\pars{\sin\pars{k\pi \over q}} $$ which yields $$ \left\lbrace\begin{array}{rcl} \ds{\Psi\pars{1 \over 3} + \gamma} & \ds{=} & \ds{-\,{\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}} \\[2mm] \ds{\Psi\pars{5 \over 6} + \gamma} & \ds{=} & \ds{{\root{3} \over 2}\,\pi - 2\ln\pars{2} - {3 \over 2}\,\ln\pars{3}} \end{array}\right. $$ and $\ds{\Psi\pars{\half} + \gamma = -2\ln\pars{2}}$.
\begin{align} \color{#f00}{I} & = \int_{0}^{\infty}\pars{2\,{1 - \expo{x} \over 1 - \expo{3x}} + {1 + \expo{x} \over 1 + \expo{3x}}}\,\dd x = \color{#f00}{\ln\pars{3}} \approx 1.0986 \end{align}

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$$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\int_{0}^{\infty}\frac{1-e^x}{1-e^{3x}}dx+2\int_{0}^{\infty}\frac{1-e^{4x}}{1-e^{6x}}dx$$ $$I=\int_{0}^{\infty}\frac{1}{e^x(1+e^x+e^{2x})}e^xdx+\int_{0}^{\infty}\frac{1+e^{2x}}{e^{2x}(1+e^{2x}+e^{4x})}2e^{2x}dx$$ Set $u=e^x$ in the first integral and $u=e^{2x}$ in the second one, so we have $$I=\int_{1}^{\infty}\frac{u+2}{u(1+u+u^2)}du=\int_{1}^{\infty}\left(\frac{2}{u}-\frac{2u+1}{1+u+u^2}\right)du=\ln\left(\frac{u^2}{1+u+u^2}\right)\Big{|}_{1}^{\infty}=\color{red}{\ln 3}$$

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  • $\begingroup$ So Thanks egreg $\endgroup$ – Behrouz Maleki Jul 18 '16 at 13:19
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By replacing $x$ with $-\log t$ we get:

$$ I = \int_{0}^{1}\frac{3t-t^2+3t^3}{1+t^2+t^4}\,dt=\int_{0}^{1}\frac{3t-t^2+t^4-3t^5}{1-t^6}\,dt\tag{1} $$ and by expanding the last integrand function as a Taylor series we get: $$ I = \sum_{n\geq 0}\left(\frac{3}{6n+2}-\frac{1}{6n+3}+\frac{1}{6n+5}-\frac{3}{6n+6}\right)\tag{2}$$ that can be easily computed through: $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{3}$$ leading to: $$\sum_{n\geq 0}\frac{1}{(6n+2)(6n+6)}=\frac{\pi\sqrt{3}+9\log 3}{144},\quad \sum_{n\geq 0}\frac{1}{(6n+3)(6n+5)}=\frac{\pi\sqrt{3}-3\log 3}{24}. \tag{4}$$

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