Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$

The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.

Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?

Noticing that
by which $$I= \left.\frac{\partial}{\partial a} I(a)\right|_{a=0} $$
where$$I(a)= \int_0^{\infty} \frac{x^a}{\left(1+x^2\right)^2} d x =\frac{1}{2} \int_0^1 y^{\frac{1-a}{2}}(1-y)^{\frac{a-1}{2}} d y\\ \quad \qquad =\frac{1}{2} B\left(\frac{3-a}{2},\frac{a+1}{2}\right) =\frac{1}{2} \Gamma\left(\frac{3-a}{2}\right) \Gamma\left(\frac{a+1}{2}\right)   $$
Differentiating both sides w.r.t. to $a$ at $a=0$ yields $$
\begin{aligned}
I=\left.\frac{\partial}{\partial a} I(a)\right|_{a=0} &=\frac{1}{2} I\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)\left(-\frac{1}{2} 4\left(\frac{3}{2}\right)+\frac{1}{2} \psi\left(\frac{1}{2}\right)\right) \\
&=-\frac{1}{8} \Gamma^2\left(\frac{1}{2}\right)\left(4\left(\frac{3}{2}\right)-4\left(\frac{1}{2}\right)\right) \\
&=-\frac{\pi}{8} \cdot 2 \\
&=-\frac{\pi}{4}
\end{aligned}
$$
