Using contour integration to solve $ \int ^\infty _0 \frac {\ln x} {(x^2+1)} dx$ Question:

Find the value of contour integration $$ \int ^{\infty}_0 \frac {\ln x} {(x^2+1)} dx$$ 

Attempt:
I just calculate 
$$\text{Res}(f,z=i) = 2\pi i\lim_{z\to i}(z - i)\frac{\ln z}{z^2+i} = \frac{\pi^2 i}2$$
Im not too sure how to move on from here. Any ideas?
 A: Consider the branch $f(z) = \frac{\ln z}{z^2 +1}$ where $|z| > 0 , -\frac{\pi}{2}< \arg z < \frac{3\pi}{2}$. Take the path $C = L_2 + L_1 + C_{\rho} + C_R$ where $\rho < 1 < R$ and $C_R$ and $C_{\rho}$ are the semi-circles with radius $R$ and $\rho$ respectively. See the figure below.
$\hskip.75in$
By Cauchy's Theorem we have 
$$\int_{L_1} f(z ) dz + \int_{L_2} f(z) \, dz  + \int_{C_\rho} f(z) \, dz + \int_{C_R} f(z) \, dz = 2\pi i \,\,\mathrm {Res}_{z = i} f(z)$$
Now if $z = re^{i\theta}$ then we may write
$$f(z) = \frac{\ln r + i\theta}{r^2e^{2i\theta} + 1}$$  
and use the parametric representations
$$z = re^{i0} = r  \,\,(\rho \leq r \leq R) \,\,\, \text{and}\,\,\,z = re^{i\pi} \,\,(\rho \leq r \leq R)$$
for the legs $L_1$ and $-L_2$ respectively, which yields 
$$\int_{L_1} f(z)\, dz - \int_{-L_2} f(z) \, dz = \int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr + \int_{\rho}^R \frac{\ln r + i\pi}{r^2 +1} \, dr$$
The residue at $z = i$ is $\mathrm {Res}_{z=i} = \frac{\pi}{4}$, then 
$$\int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr + \int_{\rho}^R \frac{\ln r + i\pi}{r^2 +1} \, dr = \frac{\pi^2i}{2} -\int_{C_\rho} f(z) \, dz - \int_{C_R} f(z) \, dz  $$
equating the real parts
$$2 \int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr = -\int_{C_\rho} f(z) \, dz - \int_{C_R} f(z) \, dz $$
It remains only to show that $\displaystyle \lim_{\rho \to 0} \int_{C_\rho} f(z) \, dz = 0 $ and $\displaystyle \lim_{R\to \infty} \int_{C_R} f(z) \, dz = 0$. 
Do you think you can take it from here?
