<s>[INST] Do we have a simpler method for computing $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x$, where $b> \frac{a^2}{4} $? Background
After finding the exact value of the integral in my post, I start to investigate a similar integral
$$I(a):
 =\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x=\int_{-\infty}^{\infty} \frac{\ln \left[\left(x+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)\right] d x}{1+x^2}$$
where $b> \frac{a^2}{4}.$

By Contour integration along the upper semi-circle
Using the fact that $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the $x^2$ to $x$ and making the branch point of $\ln$ below the real axis, we change the integral into
$$
$$
\begin{aligned}
I(a) & =2 \operatorname{Re} \int_{-\infty}^{\infty} \frac{\ln \left(x+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{1+x^2} d x \\
& =2 \operatorname{Re}\left[2 \pi i \lim _{z \rightarrow i} \frac{\ln \left(z+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{z+i}\right] \\
& =2 \operatorname{Re}\left[2 \pi i \frac{\ln \left(i+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{2 i}\right] \\
& = \pi \ln \left(1+b+\sqrt{4 b^2-a^2}\right)
\end{aligned}
For example,
$$
\begin{aligned}& \int_{-\infty}^{\infty} \frac{\ln \left(x^2+x+\frac{1}{2}   \right)}{1+x^2} d x =\pi \ln \left(\frac{5}{2}\right) \\
& \int_{-\infty}^{\infty} \frac{\ln \left(x^2+x+1\right)}{1+x^2} d x =\pi \ln (2+\sqrt{3})
\end{aligned}
$$
Do we have a simpler method for computing $$\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x,$$ where $b> \frac{a^2}{4}  $?
 [/INST] Consider, instead
\begin{align}
 &\int_{-\infty}^{\infty} \frac{\ln (x^2+2x\sqrt b \sin \theta+b)}{x^2+1} \ d x\\
=& \int_{-\infty}^{\infty}\bigg(\ln (x^2+b)+ \int_0^{\theta}\frac{2x\sqrt b \cos t}{x^2+2x\sqrt b \sin t+b} dt\bigg) \frac{dx}{x^2+1}\\
 =& \ 2\pi\ln(1+\sqrt b) -\int_0^{\theta}\frac{2\pi \sqrt b \sin t}{1+2\sqrt b \cos t+b} dt
= \overset{}{\pi}\ln \left(1+b+2\sqrt b \cos\theta\right) 
\end{align}
and then set $a=2\sqrt b \sin \theta$ to obtain
\begin{align}
 &\int_{-\infty}^{\infty} \frac{\ln (x^2+ax+b)}{x^2+1} \ d x
= \pi \ln \left(1+b+\sqrt{4 b-a^2}\right)
\end{align}
</s>