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I have trouble with this seemingly simple problem $$ K=\int_0^\infty\frac{\coth^2x-1}{(\coth x-x)^2+\frac{\pi^2}{4}}dx=\frac45.\tag{A} $$ Here is the Wolfram Alpha computation Proof.
I tried to find residue at the pole $x=\frac{\pi i}{2}$ and get $\frac{9}{5\pi i}$ (link to WA). Therefore residue theorem tell me that $K=\frac{18}{5}\neq\frac45$. I'm stuck. How to prove (A)?
To answer your question, it is not correct because, while you have calculated the residue of the integrand correctly, you are assuming that the contour of integration is a semicircle. This semicircle encompasses other poles in the complex plane, which explains why your result is not right.
Evaluating a real integral via the Residue Theorem can be a tricky business. Here's a start: what contour? The semicircle seem horrible, because how do we get the other poles? Better is a closed contour that contains only the pole at $z=i \pi/2$.
Now we need to determine the form of the integrand we need in the contour integral. Remember, somehow we should get the original integral back using a direct parameterization of the contour. So, without further ado...
Consider the contour integral
$$\oint_C dz \, \tanh{\left [z-\operatorname{arctanh}{\left (z-i \frac{\pi}{2} \right )} \right ]} $$
which is equal to
$$\oint_C dz \, \frac{\displaystyle 1-z \coth{z} + i \frac{\pi}{2} \coth{z}}{\displaystyle \coth{z}-z+i \frac{\pi}{2}} $$
where $C$ is the rectangle with vertices $\pm R$ and $\pm R+i \pi$.
The contour integral is then equal to
$$\int_{-R}^R dx \frac{\displaystyle 1-x \coth{x} + i \frac{\pi}{2} \coth{x}}{\displaystyle \coth{x}-x+i \frac{\pi}{2}} + i \int_0^{\pi} dy \, \frac{\displaystyle 1-(R+i y)\coth{(R+i y)} + i \frac{\pi}{2} \coth{(R+i y)}}{\displaystyle \coth{(R+i y)} - (R+i y) + i \frac{\pi}{2}} \\ + \int_{R}^{-R} dx \frac{\displaystyle 1-x \coth{x} - i \frac{\pi}{2} \coth{x}}{\displaystyle \coth{x}-x-i \frac{\pi}{2}}\\ + i \int_{\pi}^0 dy \, \frac{\displaystyle 1-(-R+i y)\coth{(-R+i y)} + i \frac{\pi}{2} \coth{(-R+i y)}}{\displaystyle \coth{(-R+i y)} - (-R+i y) + i \frac{\pi}{2}}$$
Now consider the second and fourth integrals, i.e., those over the vertical edges of the rectangle. We consider the limit as $R \to \infty$. Note that the integrand of the second integral approaches $1$ in this limit, while the integrand of the fourth integral approaches $-1$. (Exercise for the reader.) Thus, the sum of these two integrals is $i 2 \pi$.
We can also combine the integrand of the first and third integrals to get the integrand of the integral we seek, times $i \pi$. Thus, the contour integral is equal to (after exploiting the evenness of that integrand)
$$i 2 \pi \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} + i 2 \pi$$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i \pi/2$. Interestingly enough, this pole is not simple, but is instead a third-order pole. Given the integrand, I find it easier to compute the Laurent series directly and use that to find the residue. However, I will simply state the result as follows:
$$\operatorname*{Res}_{z=i \pi/2} \frac{\displaystyle 1-z \coth{z} + i \frac{\pi}{2} \coth{z}}{\displaystyle \coth{z}-z+i \frac{\pi}{2}} = \frac{9}{5} $$
so that
$$i 2 \pi \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} + i 2 \pi = i 2 \pi \frac{9}{5} $$
or
$$ \int_0^{\infty} dx \, \frac{\coth^2{x}-1}{\displaystyle (\coth{x}-x)^2+\frac{\pi^2}{4}} = \frac{4}{5} $$
ADDENDUM
Let's take a look at that residue calculation. For this, it helps to know that
$$\coth{\left ( z + i \frac{\pi}{2} \right )} = \tanh{z}$$
and
$$\tanh{z} = z - \frac13 z^3 + \frac{2}{15} z^5 + O \left ( z^7 \right )$$
so that the integrand looks like, in the neighborhood of $z=i \pi/2$,
$$ \frac{1-z \tanh{z}}{\tanh{z}-z} $$
Now we can find the Laurent expansion of this function about $z=0$, which looks like
$$-\frac{3}{z^3} \frac{\displaystyle 1-z^2 +O \left ( z^4 \right )}{\displaystyle 1-\frac{2}{5} z^2+O \left ( z^4 \right )} = -\frac{3}{z^3} \left [1-z^2 +O \left ( z^4 \right ) \right ] \left [1+\frac{2}{5} z^2+O \left ( z^4 \right ) \right ]$$
The residue is the coefficient of $z^2$ in the numerator, which may simply be read off as $9/5$ as stated above.
ADDENDUM II
How do we show that $z=i \pi/2$ is the only pole inside the rectangle? We can use Rouche's theorem. For example, we would just need to show that, on the rectangle,
$$\left | \coth{z}-z \pm i \frac{\pi}{2} \right | \gt |\coth{z} | $$
On the horizontal sides of the rectangle, we see that
$$\left | \coth{x}-x \pm i \frac{\pi}{2} \right |^2 - |\coth{x} |^2 = \frac{\pi^2}{4} - \left (2 x \coth{x}-x^2 \right ) $$
which is indeed $\gt 0$ for all $x \in \mathbb{R} $.
On the vertical sides of the rectangle, the inequality is obviously satisfied because $|\coth{(R + i y)}|$ approaches $1$ as $R \to \pm \infty$.
Therefore, by Rouche's theorem, the denominator of the integrand has the same number of zeroes inside the rectangle as $\coth{z}$, which is just the one at $z=i \pi/2$ and no others. Thus, the only pole inside the rectangle is at $z=i \pi/2$.
