Prove that $\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}$ 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 $I=\ln{3}$
 A: 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.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: $$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\int_{0}^{\infty}\frac{1-e^x}{1-e^{3x}}dx+2\int_{0}^{\infty}\frac{1-e^{4x}}{1-e^{6x}}dx$$
$$I=\int_{0}^{\infty}\frac{1}{e^x(1+e^x+e^{2x})}e^xdx+\int_{0}^{\infty}\frac{1+e^{2x}}{e^{2x}(1+e^{2x}+e^{4x})}2e^{2x}dx$$
Set $u=e^x$ in the first integral and $u=e^{2x}$ in the second one, so we have
$$I=\int_{1}^{\infty}\frac{u+2}{u(1+u+u^2)}du=\int_{1}^{\infty}\left(\frac{2}{u}-\frac{2u+1}{1+u+u^2}\right)du=\ln\left(\frac{u^2}{1+u+u^2}\right)\Big{|}_{1}^{\infty}=\color{red}{\ln 3}$$
A: 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:
$$\sum_{n\geq 0}\frac{1}{(6n+2)(6n+6)}=\frac{\pi\sqrt{3}+9\log 3}{144},\quad \sum_{n\geq 0}\frac{1}{(6n+3)(6n+5)}=\frac{\pi\sqrt{3}-3\log 3}{24}. \tag{4}$$
