Integral $\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx$ if $a>b>0$ I am trying to evaluate the following integral:
$$\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx,$$  where $a>b>0$.
I can't really think of a way to find it. So, please give me a hint.

With $\sin\theta =\frac ba$\begin{align}
&\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx\\
=& \int^{1}_{-1} \frac{\ln a}{x^2+1}dx+ 
 \int^{1}_{-1} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}\overset{x\to1/x}{dx}\\
=& \ \frac\pi2\ln a+\frac12 \int^{\infty}_{-\infty} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}dx-2\int_1^\infty \frac{\ln x}{x^2+1}dx\\
= &\ \frac\pi2\ln a+\frac\pi2 \ln\left(2\cos\frac\theta2\right)-2G\\
=& \ \frac\pi2\ln2 +\frac\pi2 \ln\left( \sqrt{a^2-b^2}+a\right)-2G
\end{align}
