Proving $\int_{0}^{+\infty}{(e^{-x}-2)(e^{-x}-1)\over 1- 2\cosh x}dx=\int_{0}^{+\infty}{(e^{-x}-3)(e^{-x}-2)(e^{-x}-1)\over (1- 2\cosh x)^2} dx=-1$ Proposed:

$$\int_{0}^{+\infty}{(e^{-x}-2)(e^{-x}-1)\over 1- 2\cosh x}\mathrm dx=-1\tag1$$

and 

$$\int_{0}^{+\infty}{(e^{-x}-3)(e^{-x}-2)(e^{-x}-1)\over (1- 2\cosh x)^2}\mathrm dx=-1\tag2$$

My try:
Expand $(1)$ and $(2)$
$$\int_{0}^{+\infty}{e^{-2x}-3e^{-x}+2\over 1-e^{x}-e^{-x}}\mathrm dx\tag3$$
$$\int_{0}^{+\infty}{e^{-3x}-6e^{-2x}+11e^{-x}-6\over (1-e^{x}-e^{-x})^2}\mathrm dx\tag4$$
How does one prove $(1)$ and $(2)?$
 A: $\begin{align}\int_{0}^{+\infty}{(e^{-x}-2)(e^{-x}-1)\over 1- 2\cosh x}\mathrm dx=\end{align}$
$\left[\frac{\mathrm{log}\left( {{e}^{2x}}-{{e}^{x}}+1\right) }{2}-3 \left( -x-\frac{\mathrm{arctan}\left( \frac{2{{e}^{x}}-1}{\sqrt{3}}\right) }{\sqrt{3}}+\frac{\mathrm{log}\left( {{e}^{2x}}-{{e}^{x}}+1\right) }{2}\right) +\frac{\mathrm{arctan}\left( \frac{2{{e}^{x}}-1}{\sqrt{3}}\right) }{\sqrt{3}}+\frac{4\mathrm{arctan}\left( \frac{2{{e}^{-x}}-1}{\sqrt{3}}\right) }{\sqrt{3}}+{{e}^{-x}}-x\right]_0^{+\infty}$
$=\frac{\sqrt{3}\cdot \pi }{2}+\frac{\pi }{2\cdot \sqrt{3}}-\frac{2\cdot \pi }{{{3}^{\frac{3}{2}}}}-\left(\frac{\pi }{2\cdot \sqrt{3}}+\frac{5\cdot \pi }{2\cdot {{3}^{\frac{3}{2}}}}+1\right)=-1$
A: hint
use substitution $$t=e^{-x} $$ with
$$dx=-\frac {dt}{t} $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffe]{\ds{\int_{0}^{\infty}{\pars{\expo{-x} - 2}\pars{\expo{-x} - 1} \over 1 - 2\cosh\pars{x}}\,\dd x}} =
\int_{0}^{\infty}{\pars{\expo{-x} - 2}\pars{\expo{-x} - 1} \over \expo{-2x} - 1 - \expo{-x}}\pars{-\expo{-x}}\,\dd x
\\[5mm] \stackrel{\exp\pars{-x}\ =\ t}{=}\,\,\,&\
-\int_{0}^{1}{t^{2} - 3t + 2 \over t^{2} - t + 1}\,\dd t =
\int_{0}^{1}\pars{{2t - 1 \over t^{2} - t + 1} - 1}\,\dd t =
\left.\vphantom{\Large A}\ln\pars{t^{2} - t + 1}\right\vert_{\ 0}^{\ 1} - 1 = \bbx{-1}
\end{align}

I guess the next one follows a similar trend !!!.

