Proving $\int_{\Bbb R} \frac{\cos x}{(x^2+t^2)^2}dx=\frac{\pi(t+1)}{2t^3e^t}, t>0$ 
I want to prove that $$\int_{\Bbb R}\frac{\cos x}{(x^2+t^2)^2}dx=\frac{\pi(t+1)}{2t^3e^t},t > 0$$

Integral calculator with steps gives the following answer.
$\displaystyle\int_{\Bbb R}\frac{\cos{(x)}}{(x^2+t^2)^2 }dx=\frac{(t*\cosh{(t)}-\sinh{(t)})*Si(x+it)-i*(t*\sinh{(t)}-\cosh{(t)})*Ci(x+it)+(t*\cosh{(t)}-\sinh{(t)}*Si(x-it)+i*(t*\sinh{(t)}-\cosh{(t)})*Ci(x-it)}{4*t^3}+\frac{x*\cos{(x)}}{2*t^2*(x^2+t^2)}+C$
Now my question is how to evaluate $Si{(x+it)},Ci(x+it),Si(x-it),Ci(x-it)?$
My another question is how to prove this equation using "Differentiation under Integral Sign making  change of variables"

Another way to do it is to note that$$\frac{1}{x^2+t^2}=\frac{1}{2it}\left(\frac{1}{x-it}-\frac{1}{x+it}\right)\implies\frac{1}{(x^2+t^2)^2}=\frac{-1}{4t^2}\left(\frac{1}{(x-it)^2}+\frac{1}{(x+it)^2}+\frac{i}{t}\left(\frac{1}{x-it}-\frac{1}{x+it}\right)\right).$$The $e^{ix}$ in $\cos x$ doesn't diverge when $x=i\infty$, so the $\Im x\ge0$ semicircle contains a pole $it$ contributing$$\frac{-i\pi}{2t^2}\left.\left(\frac{d}{dx}e^{ix}+\frac{i}{t}e^{ix}\right)\right|_{x=it}=\frac{-i\pi}{2t^2}\left(ie^{-t}+\frac{i}{t}e^{-t}\right)=\frac{\pi(t+1)}{2t^3e^t}.$$
