Prove that $\int_{-\infty}^\infty \frac{1}{(\coth(x)+x)^2+\frac{\pi^2}{4}} \mathrm{d}x =1 $ 
Prove that 
  $$
\int_{-\infty}^\infty \frac{1}{(\coth(x)+x)^2+\frac{\pi^2}{4}} \mathrm{d}x =1 $$

The hyperbolic function $\coth x$ is defined by:
$$
{\displaystyle \coth x={\frac {\cosh x}{\sinh x}}={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}={\frac {e^{2x}+1}{e^{2x}-1}}}.
$$
I got this problem from my friend and I tried hard to solve it but I don't know how to begin with it: all I can observe is that the denominator can never be zero. A similar problem I found in this site is this:
Prove this $ \int_0^\infty\frac{\coth^2x-1}{(\coth x-x)^2+\frac{\pi^2}{4}}dx=\frac45 $
Could anyone please help?
 A: We will adopt the standard technique (see this.2.5 or this). Let $a > 0$ and consider the function
$$ \newcommand{\Log}{\operatorname{Log}}
f(z) = \frac{1}{\frac{1}{2}\Log(-z) + a \frac{z+1}{z-1}} \cdot \frac{1}{z}, $$
where $\Log$ is the principal logarithm. Then using the substitution $x=\log\sqrt{y}$ and the keyhole contour,
\begin{align*}
\int_{-\infty}^{\infty} \frac{1}{(x + a \coth(x))^2 + \frac{\pi^2}{4}} \cdot \mathrm{d}x
&= \int_{0}^{\infty} \frac{1}{\bigl(\frac{1}{2}\log y + a \frac{y+1}{y-1} \bigr)^2 + \frac{\pi^2}{4}} \cdot \frac{\mathrm{d}y}{2y}\\
&= \frac{1}{2\pi i} \biggl( \int_{i0^{+}}^{+\infty+i0^{+}} f(z) \, \mathrm{d}z - \int_{-i0^{+}}^{+\infty-i0^{+}} f(z) \, \mathrm{d}z \biggr) \\
&= \sum_{z \in \mathbb{C}\setminus[0,\infty)} \operatorname{Res}(f, z).
\end{align*}
But it turns out that $f$ has a unique simple pole at $z = -1$ on $\mathbb{C}\setminus[0,\infty)$, and so, the above residue is easily computed as
\begin{align*}
\int_{-\infty}^{\infty} \frac{1}{(x + a \coth(x))^2 + \frac{\pi^2}{4}} \cdot \mathrm{d}x
= \frac{2}{a+1}.
\end{align*}
OP's question corresponds to the special case $a = 1$, proving that the answer is indeed $1$.
