<s>[INST] The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$   (B) $\dfrac{\pi \ln2}{3}$  (C) $\dfrac{\pi \ln2}{4}$  (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x}{x^2+4}dx = \ln x\int_0^\infty \frac{1}{x^2+4}dx-\int_0^\infty \left(\frac{d}{dx}\ln x\int_0^\infty \frac{1}{x^2+4}\right) \, dx \\[10pt]
= {} & \left[\ln x \frac{1}{2}\tan^{-1}\frac{x}{2}\right]-\int_0^\infty \frac{1}{x}\frac{1}{2}\tan^{-1}\frac{x}{2} \, dx
\end{align}
I could not move ahead. Can someone help me to get final answer?
 [/INST] In General $$\int_0^{\infty}\frac{\ln(x)}{x^2+b^2}dx=\frac{\pi\ln(b)}{2b}$$ for $b>0, b\neq 1$
$\bf{Solution::}$ Let we introduce a parameter, say $a > 0$ as follows:
Set $$F(a) =  \int_{0}^{\infty} \frac{\ln ax}{x^2 + b^2} dx$$ 
So that we get $$\frac{dF}{da} = \frac{1}{a}\int_{0}^{\infty} \frac{1}{x^2 + b^2} dx = \frac{1}{a} \frac{\pi}{2b}$$.
On integrating with respect to $a$,
We get $$\displaystyle F(a) = \frac{\pi}{2b} \ln a + C$$ and plugging in $a = 1$ gives $C = 0$
So we get $$F(a) = \int_{0}^{\infty} \frac{\ln ax}{x^2 + b^2} dx=\frac{\pi}{2b} \ln a$$
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