<s>[INST] Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$ 
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$

We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\
&=(2x+b+k)(2x+b-k).
\end{align}
Thus,
\begin{align}
I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\
&=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\
&=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right]
\end{align}
For $I_1:$ Letting $2x+b-k=t$ yields
\begin{align}
I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\
&=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t
\end{align}
Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$
If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$

It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.
 [/INST] Hint:
$$ x^2 + bx+ c^2 = (x+b/2)^2 + (c^2 - b^2/4) $$
with $c^2 - b^2/4 > 0$.
This invites change of variables using $\arctan$ as
$$d\arctan x = \frac{1}{x^2+1}dx $$
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