Integrating $\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$ without the residue theorem? Consider the integral
$$\int_{0}^{\infty}\frac{\cos(x)}{(1+x^2)^2}dx$$
Are there any other ways to compute this integral besides using the residue theorem?
Edit: Thank you all, kind people, for the answers. I had this on an exam and solved it with the residue theorem and was just wondering what other ways are possible and which of those do not involve complex integration.

Let us define integral with parameter:
\begin{align}
I(t) = \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt.
\end{align}
Taking the derivative of I:
\begin{align}
\frac{dI}{dt} = \frac{d}{dt}  \int_0 ^\infty \frac{\cos(xt)}{(1+x^2)^2}dt
=\int_0 ^\infty \frac{\partial}{\partial t} \frac{\cos(xt)}{(1+x^2)^2}dt= - \int_0^\infty \frac{x\sin(xt)}{(1+x^2)^2}dt.
\end{align}
Taking the second derivative:
\begin{align}
\frac{d^2I}{dt^2} = -\int_0^\infty \frac{x^2\cos(xt)}{(1+x^2)^2}dt =
 -\int_0^\infty \frac{(1+x^2-1)\cos(xt)}{(1+x^2)^2}dt = 
 -\int_0^\infty \frac{\cos(xt)}{(1+x^2)}dt + \int_0^\infty \frac{\cos(xt)}{(1+x^2)^2}dt
\end{align}
From here we get:
\begin{align}
I’’(t) = -\int_0^\infty \frac{\cos(tx)}{1+x^2} dt + I(t)
\end{align}
It is well known that(you can derive it similar to this metdod):
\begin{equation}
\int_0^\infty \frac{\cos(tx)}{1+x^2} dt = \frac{\pi}{2} e^{-t}
\end{equation}
So we get differential equation:
\begin{align}
I’’(t) - I(t) = -\frac{\pi}{2} e^{-t}
\end{align}
Th general solution is of the form is:
\begin{equation}
I(t) = Ae^{-t} + Be^{t}
\end{equation}
And paticular solution will be of the form $I(t) = Cte^{-t}$ and the we calculate $I’’(t) - I(t)$ to get $C = \frac{\pi}{4}$.
So final form is :
\begin{align}
I(t) = Ae^{-t} + Be^t + \frac{\pi}{4}e^{-t} t
\end{align}
To get $A$ and $B$ we calculate $I(0)$ and $I’(0)$:
\begin{align}
I(0) = \int_0^\infty \frac{1}{(1+x^2)^2} dt
\end{align}
U-sub: $u=\tan{\theta}, du = \frac{1}{\cos^2{\theta}} d\theta$
We get:
\begin{equation}
\int_0^{\frac{\pi}{2}} \frac{1}{(1+\tan^2{\theta})^2} \frac{1}{\cos^2(\theta)} dt = \int_0^{\frac{\pi}{2}} \cos^2(\theta) d\theta = \frac{\pi}{4}
\end{equation}
It is obvious that $I’(0) = 0$. After pluging in $I(t)$ and calculating $A$ and $B$ we get:
\begin{equation}
I(t) = \frac{\pi}{4}(xe^{-x} + e^{-x}).
\end{equation}
Our answer is : $I(1) = \frac{\pi}{4}(2e^{-1}) = \frac{\pi}{2e}.$
