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?

Using the result $\int_0^{\infty} \frac{\ln x}{x^2+1} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x &=\frac{1}{a} \int_0^{\infty} \frac{\ln a+\ln y}{y^2+1} d y \\
&=\frac{\ln a}{a} \int_0^{\infty} \frac{d y}{y^2+1}+\frac{1}{a} \int_0^{\infty} \frac{\ln y}{y^2+1} d y \\
&=\frac{\pi \ln a}{2 a}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ yields a general integral
$$
\boxed{\int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2}=\frac{\pi}{4 a^3}(\ln a-1)}
$$
Putting $a=1$ gives our integral $$
\int_0^{\infty} \frac{\ln x}{\left(x^2+1\right)^2} d x=-\frac{\pi}{4}
$$
