<s>[INST] 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?
 [/INST] Here is an easy way to compute it :
Let's substitute $ \left\lbrace\begin{matrix}y=\frac{1}{x}\ \ \\ \mathrm{d}x=-\frac{\mathrm{d}y}{y^{2}}\end{matrix}\right. $, we get : \begin{aligned}\int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\int_{0}^{+\infty}{\frac{y^{2}\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}\\ &=\left[\frac{y\ln{y}}{2\left(1+y^{2}\right)}\right]_{0}^{+\infty}-\frac{1}{2}\int_{0}^{+\infty}{\frac{1+\ln{y}}{1+y^{2}}\,\mathrm{d}y}\\ &=-\frac{1}{2}\int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}-\frac{1}{2}\int_{0}^{+\infty}{\frac{\mathrm{d}y}{1+y^{2}}}\\ &=-\frac{1}{2}\left[\arctan{y}\right]_{0}^{+\infty}\\ \int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\frac{\pi}{4}\end{aligned}
$\textbf{Note :}$ We first integrated by parts, setting $ u':y\mapsto -\frac{y}{\left(1+y^{2}\right)^{2}} $, and $ v:y\mapsto y\ln{y} $. From the third to the fourth line, we used the fact that $ \int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}=0 $, which can be proven substituting $ y=\frac{1}{x} $.
$\textbf{Bonus :}$ Using the same method, prove that $ \left(\forall x>0 \right) $, we have : $$ \int_{\frac{1}{x}}^{x}{\frac{\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}\left(\frac{\pi}{2}-2\arctan{x}\right)+\frac{x\ln{x}}{1+x^{2}} $$
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