<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] $\bf{My\; Solution::}$ Let  $\displaystyle I = \int_{0}^{\infty}\frac{\ln(x)}{x^2+4}dx\;,$ Now Let $x=2t\;,$ Then $dx = 2dt$
and Changing Limits, we get
$$\displaystyle I = \int_{0}^{\infty}\frac{\ln(2t)}{4t^2+4}\times 2dt = \frac{1}{2}\int_{0}^{\infty}\frac{\ln(2t)}{t^2+1}dt=\frac{1}{2}\int_{0}^{\infty}\frac{\ln(2)}{t^2+1}dt+\frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{t^2+1}dt$$
So $$\displaystyle I = \frac{\ln(2)}{2}\int_{0}^{\infty}\frac{1}{1+t^2}dt+\frac{1}{2}J = \frac{\ln(2)}{2}\cdot \frac{\pi}{2}+\frac{1}{2}J...........(1)$$
Now For Calculation of $$\displaystyle J = \frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{t^2+1}dt\;,$$ Put $\displaystyle t=\frac{1}{u}$ and $\displaystyle dt = -\frac{1}{u^2}du$
and Changing Limits, We get $$\displaystyle J=-\frac{1}{2}\int_{0}^{\infty}\frac{\ln(u)}{1+u^2}du = -\frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{1+t^2}dt = -J$$
So We get $J = -J\Rightarrow J=0\;,$ Now Put into $(1)$ equation, We get $$\displaystyle I=\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}dx = \frac{\pi\cdot \ln(2)}{4}.$$
</s>