Evaluate logarithmic integral $\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ I need help to evaluate the integral with the residue theorem:
$$\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$$
where a,b>0 real numbers.
I think I could consider the contour integral where C is the half circle in the first two quadrant. 
But I'm not sure how to continue.
Could someone help me?

Let $I(a)=2\int_{0}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ and evaluate
$$I’(a) =\int_{0}^{\infty}\frac{4adx}{(x^2+a^2)(x^2+b^2)}= 
\frac{2\pi}{b(a+b)}$$
Then,
\begin{align}
\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx
& = I(a)=I(0)+ \int_{0}^{a}I’(t)dt\\
&=2 \int_{0}^{\infty} \frac{\ln{x^2}}{x^2+b^2}\,dx
 + \frac{2\pi}b  \int_{0}^{a}\frac1{t+b}dt\\
 &= \frac{2\pi }b \ln b + \frac{2\pi }b \ln \frac{a+b}b= \frac{2\pi }b \ln (a+b)
\end{align}
