How to evaluate $ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $ Evaluate $$  \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $$ $$(a>0) $$
How can I use contour appropriately?
What is the meaning of this integral?

(additionally posted)
I tried to solve this problem.
First, I take a branch $$  \Omega=\mathbb C - \{z|\text{Re}(z)=0\; \text{and} \; \text{Im}(z)\le0\}  $$ 
Then ${\log_\Omega z}=\log r +i\theta (-\frac{\pi}{2}\lt\theta\lt\frac{3\pi}{2})$
Now, $\frac{\log z}{(z^2+a^2)^2}$ is holomorphic in $\Omega - \{ai\}$ with double poles at $ai$.
Now I'll take the contour which forms an indented semicircle.
For any $0\lt\epsilon\lt{a}$, where $\max (1,a)\lt R$, $\Gamma_{R,\epsilon}\subseteq\Omega - \{ai\}$ and in $\Omega$, $i=e^{i\pi/2}$.
Now using the residue formula, $$2\pi{i}\operatorname*{Res}_{z=ai}\frac{\log_\Omega{z}}{(z^2+a^2)^2}=2\pi{i}\operatorname*{lim}_{z\to ai}\frac{d}{dz}(z-ai)^2\frac{\log_\Omega{z}}{(z^2+a^2)^2}=\frac{\pi}{2a^3}(\log_\Omega{ai}-1)$$
Now, the last part, take $i=e^{i\pi/2}$, then is equal to $\frac{\pi}{2a^3}(\log{a}-1+i\pi/2)$
So, I can split integrals by four parts, 
$$\int_{\epsilon}^R dz + \int_{\Gamma_R} dz + \int_{-R}^{-\epsilon} dz + \int_{\Gamma_\epsilon} dz$$
First, evaluate the second part,
$$\left|\int_{\Gamma_R} dz\right|\le\int_0^{\pi}\left|\frac{\log_\Omega{Re^{i\theta}}}{(R^2e^{2i\theta}+a^2)^2}iRe^{i\theta}\right|d\theta$$
Note that
$$\left|\log_\Omega{Re^{i\theta}}\right|=\left|\log R+i\theta\right|\le\left|\log R\right|+|\theta|$$
$$\left|R^2e^{2i\theta}+a^2\right|\ge R^2-a^2\quad (R\gt a)$$
Then, 2nd part $\le\frac{R(\pi R+\frac{\pi^2}{2})}{(R^2+a^2)^2}\to 0\; \text{as} \; R \to \infty\quad \left|\log R\right|\lt R\;\text{where}\;(R\gt 1)$
So, 4th part similarly, goes to $\;0$.
Then 3rd part, substitute for $\;t=-z$,
$$\int_\epsilon^{R}\frac{\log t}{(t^2+a^2)^2}dt + i\pi\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}$$
And $\;i\pi\lim\limits_{{\epsilon \to 0},\;{R\to\infty}}\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}=\frac{\pi}{4a^3}$
With tedious calculations, I got $\frac{\pi}{4a^3}(\log a -1)$.

Real Method
We first evaluate the integral
$$
J=\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x
$$
Putting $x=\tan \theta$ yields
$$
\begin{aligned}
J& =\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\ln (a \tan \theta)}{a^2 \sec ^2 \theta} \cdot a \sec ^2 \theta d \theta \\
& =\frac{1}{a} \int_0^{\frac{\pi}{2}} \ln (a \tan \theta)  d \theta \\
& =\frac{1}{a} \int_0^{\frac{\pi}{2}} \ln a d \theta+  \underbrace{\int_0^{\frac{\pi}{2}} \ln (\tan \theta) d \theta }_{=0}  \\
& =\frac{\pi}{2} \frac{\ln a}{a}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ yields
$$
\begin{aligned}
& -2 a \int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2} d x=\frac{\pi}{2} \frac{1-\ln a}{a^2} \\
\Rightarrow & \int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2}=\frac{\pi(\ln a-1)}{4 a^3}
\end{aligned}
$$
