Integrals of the form
$$\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx,$$
where $p$ is a polynomial can be evaluated by shifting the contour of integration to a line $\operatorname{Im} z \equiv c$. We first check that the integrals over the vertical segments connecting the two lines tend to $0$ as the real part tends to $\pm\infty$:
$$\lvert \cosh (x+iy)\rvert^2 = \lvert \cosh x\cos y + i \sinh x\sin y\rvert^2 = \sinh^2 x + \cos^2 y,$$
so the integrand decays exponentially and
$$\left\lvert \int_{R}^{R + ic} \frac{p(z)}{\cosh z}\,dz\right\rvert
\leqslant \frac{K\,c}{\sinh R}\left(R^2+c^2\right)^{\deg p/2} \xrightarrow{R\to \pm\infty} 0.$$
Since $\cosh \left(z+\pi i\right) = -\cosh z$, and the only singularity of the integrand between $\mathbb{R}$ and $\mathbb{R}+\pi i$ is a simple pole at $\frac{\pi i}{2}$ (unless $p$ has a zero there, but then we can regard it as a simple pole with residue $0$) with the residue
$$\operatorname{Res}\left(\frac{p(z)}{\cosh z};\, \frac{\pi i}{2}\right) = \frac{p\left(\frac{\pi i}{2}\right)}{\cosh' \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{\sinh \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{i},$$
the residue theorem yields
$$\begin{align}
\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx
&= 2\pi\, p\left(\frac{\pi i}{2}\right) + \int_{\pi i-\infty}^{\pi i+\infty} \frac{p(z)}{\cosh z}\,dz\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \int_{-\infty}^\infty \frac{p(x+\pi i)}{\cosh x}\,dx\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \sum_{k=0}^{\deg p} \frac{(\pi i)^k}{k!}\int_{-\infty}^\infty \frac{p^{(k)}(x)}{\cosh x}\,dx.\tag{1}
\end{align}$$
Since $\cosh$ is even, only even powers of $x$ contribute to the integrals, hence we can from the beginning assume that $p$ is an even polynomial, and need only consider the derivatives of even order.
For a constant polynomial, $(1)$ yields
$$\int_{-\infty}^\infty \frac{dx}{\cosh x} = 2\pi - \int_{-\infty}^\infty \frac{dx}{\cosh x}\Rightarrow \int_{-\infty}^\infty \frac{dx}{\cosh x} = \pi.$$
For $p(z) = z^2$, we obtain
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx &= 2\pi \left(\frac{\pi i}{2}\right)^2 - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx - (\pi i)^2\int_{-\infty}^\infty \frac{dx}{\cosh x}\\
&= - \frac{\pi^3}{2} - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx + \pi^3,
\end{align}$$
which becomes
$$\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx = \frac{\pi^3}{4}.$$
Partial solution
We can present the second integral as a series (imaginary part of the integral forms a series)
$$I= \int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx=\int_{0}^{\infty} \frac{(x \text{coth} (x) - 1)(2\pi-ix)}{x^2 ((2 \pi)^2 + x^2)} \, dx$$
$$x\coth x=1+2\sum_{k=1}^{\infty}\frac{x^2}{\pi^2k^2+x^2}$$
$$I=2\int_{0}^{\infty}dx\frac{2\pi-ix}{(2\pi)^2+x^2}\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2+x^2}=\Re I+i\Im I=I_1+I_2$$
$$I_1=2\pi\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (4 \pi^2 + x^2)}dx=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\frac{1}{\pi^2k^2+x^2}dx$$$$=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{dx}{\pi^2k^2-4\pi^2}$$
$$=4\pi\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2-4\pi^2}\Bigl(\frac{1}{2\pi}-\frac{1}{\pi k}\Bigr)\frac{\pi}{2}=2\pi^2\sum_{k=1}^{\infty}\frac{1}{2\pi^2k(2\pi+\pi k)}=\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$$
$$I_1=\frac{1}{2\pi}\sum_{k=1}^{\infty}\biggl(\frac{1}{k}-\frac{1}{k+2}\biggr)=\frac{3}{4\pi}$$
$$I_2=-2i\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\,\frac{x}{\pi^2k^2+x^2}dx$$
For $k\neq2$
$$I_2=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{xdx}{\pi^2k^2-4\pi^2}$$
$$=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{d(x^2)}{(2\pi)^2(1+\frac{x^2}{(2\pi)^2})}-\frac{d(x^2)}{(\pi k)^2(1+\frac{x^2}{(\pi k)^2})}\Bigr)\frac{1}{\pi^2k^2-4\pi^2}$$ $$=-\frac{4i}{\pi^2}\sum_{k\neq2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$
For $k=2$
$$I_2=-2i\int_{0}^{\infty}\frac{xdx}{((2\pi)^2+x^2)^2}=-i\int_{0}^{\infty}\frac{dt}{((2\pi)^2+t)^2}=-\frac{i}{4\pi^2}$$
$$I=I_1+I_2=\frac{3}{4\pi}-\frac{4i}{3\pi^2}\log2-\frac{i}{4\pi^2}-\frac{4i}{\pi^2}\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$
Best Answer
As shown in gimusi's answer, you do not need to compute the integral.
However, if you want to integrate, consider $$I=\int_{0}^{p}\left(\frac{1}{\sqrt{x^2+4}}-\frac{k}{x+2}\right)\,\text dx=\sinh ^{-1}\left(\frac{p}{2}\right)-k \log (p+2)+k \log (2)$$ and use series expansion for large $p$. This should give $$I=(1-k) \log \left({p}\right)+k \log (2)-\frac{2 k}{p}+\frac{2 k+1}{p^2}+O\left(\frac{1}{p^3}\right)$$ and look at the limit when $p\to \infty$.