<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] Contour Integration Approach
Consider the keyhole contour, $\gamma$, that goes from $0$ to $\infty$ just above the positive real axis, circles the complex plane counterclockwise, then returns to $0$ just below the positive real axis.
Now compute the integral
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_\gamma\frac{\log(z)^2}{\left(1+z^2\right)^2}\,\mathrm{d}z
&=\int_0^\infty\frac{\log(x)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x
-\int_0^\infty\frac{(\log(x)+2\pi i)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag1\\
&=-4\pi i\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x
+4\pi^2\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x\tag2\\
&=2\pi i\left(\color{#C00}{\Res_{z=i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}+\color{#090}{\Res_{z=-i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}\right)\tag3\\
&=2\pi i\left(\color{#C00}{-\frac\pi4+i\frac{\pi^2}{16}}+\color{#090}{\frac{3\pi}4-i\frac{9\pi^2}{16}}\right)\tag4\\[9pt]
&=i\pi^2+\pi^3\tag5
\end{align}
$$
Explanation:
$(1)$: integrate along the contour above and below the positive real axis
$\phantom{\text{(1):}}$ the integral along the huge circle vanishes
$\phantom{\text{(1):}}$ the integral near zero also vanishes
$(2)$: expand $(\log(x)+2\pi i)^2$ and combine
$(3)$: the original integral is $2\pi i$ times the sum of the residues of the integrand
$(4)$: compute the residues
$(5)$: simplify
Thus, we get not only that
$$
\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x=-\frac\pi4\tag6
$$
but also that
$$
\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x=\frac\pi4\tag7
$$

A Real Approach
This is similar to what is attempted in the question.
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^2+a^2}\,\mathrm{d}x
&=\frac1a\int_0^\infty\frac{\log(x)+\log(a)}{x^2+1}\,\mathrm{d}x\tag8\\
&=\frac{\log(a)}a\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x\tag9\\
&=\frac{\pi\log(a)}{2a}\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}\text{(8)}$: substitute $x\mapsto ax$
$\phantom{1}\text{(9)}$: substituting $x\mapsto1/x$ shows that $\int_0^\infty\frac{\log(x)}{x^2+1}\,\mathrm{d}x=0$
$(10)$: evaluate the integral
Taking the derivative of $(10)$ in $a$ and dividing by $-2a$ gives
$$
\int_0^\infty\frac{\log(x)}{\left(x^2+a^2\right)^2}\,\mathrm{d}x=\frac\pi4\frac{\log(a)-1}{a^3}\tag{11}
$$
Plug in $a=1$
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