TITLE: How may we show that $\int_{0}^{1}{1-x\over 1+x}\cdot{2k+3+x^2\over 1+x^2}\cdot{\mathrm dx\over \ln x}=-\ln(2^k\pi)?$
QUESTION [7 upvotes]: A simple closed form

$$\int_{0}^{1}{1-x\over 1+x}\cdot{2k+3+x^2\over 1+x^2}\cdot{\mathrm dx\over \ln x}=-\ln(2^k\pi)\tag1$$
  Where $k$ is a real number

Making an attempt:
This integral is too difficult to even try to make an attempt. 
Anyway let's try $x=e^y\implies dx=e^ydy$, then $(1)$ becomes
$$-\int_{0}^{\infty}{1-e^{-y}\over 1+e^{-y}}\cdot{2k+3+e^{-2y}\over 1+e^{-2y}}\cdot{e^{-y}\over y}\mathrm dy\tag2$$
Using hyperbolic identities 
$$-\int_{0}^{\infty}\tanh\left({y\over 2}\right)\cdot{2k+3+e^{-2y}\over 1+e^{-2y}}\cdot{e^{-y}\over y}\mathrm dy\tag3$$
$$-\int_{0}^{\infty}\tanh\left({y\over 2}\right)\cdot\left[(k+1)e^{2y}sech{y}+1\right]\cdot{e^{-y}\over y}\mathrm dy\tag4$$
Another try: $x=\tan y\implies dx=\sec^2y dy$, then $(1)$ becomes
$$\int_{0}^{\pi/4}\tan\left({\pi\over 4}-y\right)\cdot{2k+2+\sec^2y\over \ln\tan y}\mathrm dy\tag5$$
Still going no where!
How can we prove $(1)?$

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}}
\,{\dd x \over \ln\pars{x}} =
-\ln\pars{2^{k}\pi}:\ {\large ?}.\qquad k \in \mathbb{R}}$.

\begin{align}
&\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}}
\,{\dd x \over \ln\pars{x}} =
-\int_{0}^{1}{1 \over 1 + x}\,\pars{1 + 2\,{k + 1 \over 1 + x^{2}}}
\,{x - 1 \over \ln\pars{x}}\,\dd x
\\[5mm] = &\
-\int_{0}^{1}{1 \over 1 + x}\,{x - 1 \over \ln\pars{x}}\,\dd x -
2\pars{k + 1}\int_{0}^{1}{1 \over 1 + x}\,{1 \over 1 + x^{2}}
\,{x - 1 \over \ln\pars{x}}\,\dd x
\\[5mm] = &\
-\int_{0}^{1}{1 - x \over 1 - x^{2}}\,{x - 1 \over \ln\pars{x}}\,\dd x -
2\pars{k + 1}\int_{0}^{1}{1 - x \over 1 - x^{4}}
\,{x - 1 \over \ln\pars{x}}\,\dd x =
\bbx{\ds{a_{1} + 2\pars{k + 1}a_{2}}}
\label{1}\tag{1}
\\[5mm] &\
\mbox{where}\quad
a_{n} \equiv
-\int_{0}^{1}{1 - x \over 1 - x^{2n}}\,{x - 1 \over \ln\pars{x}}\,\dd x
\label{2}\tag{2}
\end{align}

With the identity
  $\ds{{x - 1 \over \ln\pars{x}} = \int_{0}^{1}x^{t}\,\dd t}$, $\ds{a_{n}}$ becomes:

\begin{align}
a_{n} & =
\int_{0}^{1}\int_{0}^{1}{x^{t + 1} - x^{t} \over 1 - x^{2n}}\,\dd x\,\dd t
\,\,\,\stackrel{x^{2n}\ \mapsto\ x}{=}\,\,\,
{1 \over 2n}
\int_{0}^{1}\int_{0}^{1}
{x^{t/\pars{2n}\ +\ 1/n\ -\ 1}\ -\ x^{t/\pars{2n}\ +\ 1/\pars{2n}\ -\ 1} \over
1 - x}\,\dd x\,\dd t
\\[5mm] = &\
{1 \over 2n}\int_{0}^{1}\bracks{%
\Psi\pars{t + 1 \over 2n} - \Psi\pars{t + 2 \over 2n}}\,\dd t
\end{align}
Here, I used Digamma function identity $\ds{\mathbf{6.3.22}}$ in A & S Table. With
$\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$:
\begin{align}
a_{n} & =
\left.\ln\pars{\Gamma\pars{\bracks{t + 1}/\bracks{2n}} \over
\Gamma\pars{\bracks{t + 2}/\bracks{2n}}}\right\vert_{\ t\ =\ 0}^{\ t\ =\ 1} =
\ln\pars{{\Gamma\pars{1/n} \over \Gamma\pars{3/\bracks{2n}}}\,
{\Gamma\pars{1/n} \over \Gamma\pars{1/\bracks{2n}}}}
\\[5mm] = &\
\ln\pars{\Gamma^{2}\pars{1/n} \over \Gamma\pars{3/\bracks{2n}}\Gamma\pars{1/\bracks{2n}}}
\\[5mm] 
\mbox{and} &\ 
\left\{\begin{array}{rcl}
\ds{a_{1}} & \ds{=} &
\ds{\ln\pars{\Gamma^{2}\pars{1} \over \Gamma\pars{3/2}\Gamma\pars{1/2}} =
-\ln\pars{\pi \over 2}}
\\[2mm]
\ds{a_{2}} & \ds{=} &
\ds{\ln\pars{\Gamma^{2}\pars{1/2} \over \Gamma\pars{3/4}\Gamma\pars{1/4}} =
-\ln\pars{\pi \over \pi/\sin\pars{\pi/4}} = -\,{1 \over 2}\,\ln\pars{2}}
\end{array}\right.\label{3}\tag{3}
\end{align}

With \eqref{1} and \eqref{3}:

\begin{align}
&\int_{0}^{1}{1 - x \over 1 + x}\,{2k + 3 + x^{2} \over 1 + x^{2}}
\,{\dd x \over \ln\pars{x}} =
-\ln\pars{\pi \over 2} + 2\pars{k + 1}\bracks{-\,{1 \over 2}\,\ln\pars{2}} =
\bbx{\ds{-\ln\pars{2^{k}\pi}}}
\end{align}


Note that
  $\ds{\Gamma\pars{1} = 1\,,\ \Gamma\pars{1 \over 2} = \root{\pi}}$ and the use of $\ds{\Gamma}$-Recurrence Property and Euler Reflection Formula.