Is this definite integral possible to evaluate? $\int_0^\infty\frac{\log x}{(1+x^2)^2}\,dx$ 
$$\int_0^\infty\frac{\log x}{(1+x^2)^2}\,dx$$

I have tried to evaluate this by trying I.B.P. on various different products, and have also tried the substitution $x\mapsto1/x$. None of these work. I have put this into an online integration tool and it gives a solution using some function called "Li", but I have never been taught this, so do not want to use it. Is there some way of evaluating this definite integral, perhaps using complex analysis and contour integrals?

I have been able to evaluate similar integrals, such as $\int_0^\infty\frac{\log x}{1+x^2}\,dx=0$ and $\int_0^\infty\frac{1}{(1+x^2)^2}\,dx=\pi/4$, but it is this one which is proving to be a struggle.
 A: 
Try the following integration by parts:
$$\begin{align}
\int_{0}^{\infty}\frac{\ln{\left(x\right)}}{\left(1+x^{2}\right)^{2}}\,\mathrm{d}x
&=\int_{0}^{\infty}\frac{2x}{\left(1+x^{2}\right)^{2}}\cdot\frac{\ln{\left(x\right)}}{2x}\,\mathrm{d}x\\
&=\left[\left(1-\frac{1}{1+x^{2}}\right)\frac{\ln{\left(x\right)}}{2x}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(1-\frac{1}{1+x^{2}}\right)\frac{1-\ln{\left(x\right)}}{2x^{2}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\
&=0-0+\int_{0}^{\infty}\frac{x^{2}}{1+x^{2}}\cdot\frac{\ln{\left(x\right)}-1}{2x^{2}}\,\mathrm{d}x\\
&=\frac12\int_{0}^{\infty}\frac{\ln{\left(x\right)}-1}{1+x^{2}}\,\mathrm{d}x.\\
\end{align}$$
Incidentally, the same method can be used to build a reduction formula for integrals of the form $\int_{0}^{\infty}\frac{\ln{\left(x\right)}}{\left(1+x^{2}\right)^{n}}\,\mathrm{d}x$.
A: Change variable to $y = \frac1x$, we have
$$I = \int_0^\infty \frac{\log x}{(1+x^2)^2} dx
= \int_\infty^0 \frac{-\log y}{(1+y^{-2})^2}\left(-\frac{dy}{y^2}\right)
= - \int_0^\infty \frac{y^2\log y}{(1+y^2)^2}dy$$
This leads to
$$\begin{align}
I &= -\frac12\int_0^\infty \log(x) \frac{x^2-1}{(x^2+1)^2}dx
= -\frac12\int_0^\infty\log(x) \frac{x-x^{-1}}{(x+x^{-1})^2}\frac{dx}{x}\\
&= -\frac12\int_0^\infty\log(x)\frac{x-x^{-1}}{(x+x^{-1})^2}\frac{d(x+x^{-1})}{x-x^{-1}}
= -\frac12\int_0^\infty \log(x) \frac{d(x+x^{-1})}{(x+x^{-1})^2}\\
&= \frac12\int_0^\infty \log(x) \left(\frac{1}{x+x^{-1}}\right)' dx
= \frac12\int_0^\infty \left[ \left(\frac{\log x}{x+x^{-1}}\right)' - \frac{(\log x)'}{x+x^{-1}} \right] dx\\
&= \left[ \frac{\log x}{x+x^{-1}}\right]_0^\infty -\frac12\int_0^\infty \frac{dx}{1+x^2}\\
&= -\frac{\pi}{4}
\end{align}
$$
A: Take the keyhole contour of $f(z)=\left[\frac{\ln(z)}{1+z^2}\right]^2$ with $\gamma$ the radius of the inner circle and $\Gamma$ the radius of the outer circle to get
$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_1}f(z)\ dz=\int_0^{+\infty}\left[\frac{\ln(x)}{1+x^2}\right]^2\ dx$$
$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_3}f(z)\ dz=-\int_0^{+\infty}\left[\frac{\ln(x)+2\pi i}{1+x^2}\right]^2\ dx$$
$$\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_2}f(z)\ dz=\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\int_{C_4}f(z)\ dz=0$$
And thus, we can see that
$$\begin{align}\lim_{(\gamma,\Gamma)\to(0^+,+\infty)}\oint_Cf(z)\ dz&=\int_0^{+\infty}\left[\frac{\ln(x)}{1+x^2}\right]^2\ dx-\int_0^{+\infty}\left[\frac{\ln(x)+2\pi i}{1+x^2}\right]^2\ dx\\&=\int_0^{+\infty}\frac{[\ln(x)]^2-[\ln(x)+2\pi i]^2}{(1+x^2)^2}\ dx\\&=\int_0^{+\infty}\frac{[\ln(x)]^2-[\ln(x)]^2-4\pi i\ln(x)+4\pi^2}{(1+x^2)^2}\ dx\\&=\int_0^{+\infty}\frac{-4\pi i\ln(x)+4\pi^2}{(1+x^2)^2}\ dx\\&=2\pi i\left(\operatorname{Res}_{z=i}(f(z))+\operatorname{Res}_{z=-i}(f(z))\right)\\&=\pi^2i\end{align}$$
By taking imaginary parts, we deduce that
$$\int_0^{+\infty}\frac{\ln(x)}{(1+x^2)^2}\ dx=-\frac\pi4$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x & =
\left.-\,
\partiald{}{a}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + a}\,\dd x
\,\right\vert_{\ a=\ 1}
\\[5mm] & =
-\,\partiald{}{a}
\bracks{a^{-1/2}\int_{0}^{\infty}{\ln\pars{a^{1/2}x} \over x^{2} + 1}\,\dd x}
_{\ a=\ 1}
\\[5mm] & =
-\,\partiald{}{a}
\bracks{{1 \over 2}\,a^{-1/2}\ln\pars{a}\int_{0}^{\infty}{\dd x \over x^{2} + 1} +
a^{-1/2}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x}\label{1}\tag{1}
_{\ a=\ 1}
\\[5mm] & =
-\,{\pi \over 4}\,\partiald{}{a}
\bracks{a^{-1/2}\ln\pars{a}}_{\ a\ =\ 1}
\\[5mm] & =
-\,{\pi \over 4}
\bracks{-\,{1 \over 2}\,a^{-3/2}\ln\pars{a} + a^{-3/2}}_{\ a\ =\ 1} =
\bbx{-\,{\pi \over 4}}
\end{align}

The last integral in \eqref{1} vanishes out: It just changes its sign under $\ds{x \to 1/x}$.

A: The substitution $x= \frac{1}{y}$ on the interval $(1,\infty)$ followed by the substitution $y=\tan(t)$ and integration by parts yield
\begin{gather*}
\int_{0}^{\infty}\dfrac{\ln(x)}{(1+x^2)^2}\, dx = \int_{0}^{1}\dfrac{\ln(x)}{(1+x^2)^2}\, dx - \int_{0}^{1}\dfrac{y^2\ln(y)}{(1+y^2)^2}\, dy = \int_{0}^{1}\dfrac{(1-y^2)\ln(y)}{(1+y^2)^2}\, dy = \\[2ex] \int_{0}^{\pi/4}\cos(2t)\ln(\tan(t))\, dt = \left[\dfrac{1}{2}\sin(2t)\ln(\tan(t))\right]_{0}^{\pi/4}- \int_{0}^{\pi/4}\underbrace{\dfrac{\sin(2t)}{2\tan(t)\cos^2(t)}}_{=1}\, dt = -\dfrac{\pi}{4}.
\end{gather*}
