$\int_0^\infty \frac{\log(1+x^2)}{x^2} dx $ using contour integration

Notice that by integrating by parts we have: $$ I = \int_{0}^{+\infty}\frac{\log(1+x^2)}{x^2}\,dx = -\left.\frac{1}{x}\log(1+x^2)\right|_{0}^{+\infty}+\int_{0}^{+\infty}\frac{2\,dx}{1+x^2}=\color{red}{\int_{-\infty}^{+\infty}\frac{dz}{1+z^2}} $$ and the last integral is very easy to evaluate with complex or real-analytic techniques.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Note that \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{2}}\,\dd x} =\half\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{3/2}}\,\dd x =\half\int_{1}^{\infty}\pars{x - 1}^{-3/2}\,\,\ln\pars{x}\,\dd x \qquad\pars{1} \end{align} We evaluate the integral, in the RHS, by performing the contour integral $$ \oint_{\gamma}\pars{z - 1}^{-3/2}\,\,\ln\pars{z}\,\dd z $$ along the path $\ds{\gamma}$ depicted below:

which shows two branch cuts. Namely: \begin{align} \begin{array}{rclrcll} \pars{z - 1}^{-3/2} & = &\verts{z - 1}^{-3/2} \exp\pars{-\,{3 \over 2}\,\,{\rm Arg}\pars{z - 1}\ic}\,, & z & \not= & 1 \,, & \phantom{-}0 < \,{\rm Arg}\pars{z - 1} < 2\pi \\[5mm] \ln\pars{z}&=&\ln\pars{\verts{z}} + \ic\,{\rm Arg}\pars{z}\,,& z &\not = & 0\,, & -\pi < \,{\rm Arg}\pars{z} < \pi \end{array} \end{align} Obviously, the integral along $\ds{\gamma}$ vanishes out because there are not any poles inside the contour. For simplicity, we omit the contribution of the arcs $\ds{C_{R}}$, of radius $\ds{R}$, and the 'small' semicircles, of radius $\ds{\epsilon}$, around $\ds{z=0}$ and $\ds{z=1}$: They don't yield any contribution in the limits $\ds{R\ \to\ \infty\,,\ \epsilon\ \to\ 0^{+}}$. Then, \begin{align} 0&=\oint_{\gamma}\pars{z - 1}^{-3/2}\,\,\ln\pars{z}\,\dd z =\overbrace{\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ln\pars{x}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{++}}}} \\[5mm]&+\overbrace{% \int_{-\infty}^{0}\pars{1 - x}^{-3/2}\expo{-3\pi\ic/2} \bracks{\ln\pars{x} + \ic\pi}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{-+}}}}\ +\ \overbrace{% \int^{-\infty}_{0}\pars{1 - x}^{-3/2}\expo{-3\pi\ic/2} \bracks{\ln\pars{x} - \ic\pi}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{--}}}} \\[5mm]&+\overbrace{\int^{1}_{\infty}\pars{x - 1}^{-3/2} \expo{-3\pi\ic}\ln\pars{x}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{+-}}}}\ =\ 2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ln\pars{x}\,\dd x \\[5mm]&\phantom{=}+2\pi\ic\expo{-3\pi\ic/2}\int_{-\infty}^{0} \pars{1 - x}^{-3/2}\,\dd x =2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x -\left.{4\pi \over \root{1 - x}}\,\right\vert_{\, x\ \to\ -\infty}^{\, x\ =\ 0} \\[5mm]&=2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x - 4\pi \quad\imp\quad \boxed{\ds{\quad\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x =2\pi\quad}} \end{align}

Replacing this result in expression $\pars{1}$, we'll find:

\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{2}}\,\dd x} =\color{#66f}{\Large\pi} \end{align}

Since $\frac{log(1+x^2)}{x^2} = \lim_{y \to 0} \frac{log(1+x^2)}{x^{2}+y^2}$ almost everywhere, and the latter function is equal to $2\, \Re \frac{log(1-ix)}{x^{2}+y^{2}}$ on the real line, we may write the integral as $$\Re\lim_{y \to 0}\int_{-\infty}^{\infty} \! \frac{log(1-ix)}{x^{2}+y^{2}} \, \mathrm{d}x.$$ To avoid the branch cut, the contour is closed above the real line (usual D-shape) and we get the result $$\Re\lim_{y \to 0}2 \pi i Res_{x=iy}\frac{log(1-ix)}{x^{2}+y^{2}}=\Re\lim_{y \to 0}2 \pi i \frac{log(1+y)}{2 i y}=\pi.$$

Let $f(z)=\dfrac{\log\left(1+z^2\right)\log z}{z^2}$ and integrate it along a positively-oriented, triply-indented circular contour $C$ that avoids branch cuts taken at $\pm i[1,\infty)$ and $[0,\infty)$. Essentially the same setup as used in this answer, just ignore the pole/white spot at $-1$. Note that the singularity at $z=0$ is removable and $C$ contains no poles, so $\displaystyle\oint_C f(z)\,dz = 0$.

As the large arc gets larger, and the arcs enclosing the branch points get smaller, the overall integral over $C$ will converge to the total of the integrals to either side of each cut, and the result follows:

$$\begin{align*} 0 &= \int_0^\infty \frac{\log\left(1+x^2\right)\log x}{x^2} \, dx + \int_\infty^0 \frac{\log\left(1+x^2\right) \left(\log x+i2\pi\right)}{x^2} \, dx \\ &\quad + i \int_\infty^1 \frac{\left(\log\left\lvert1-x^2\right\rvert+i\pi\right) \left(\log x+i\frac\pi2\right)}{-x^2} \,dx + i \int_1^\infty \frac{\left(\log\left\lvert1-x^2\right\rvert-i\pi\right) \left(\log x+i\frac\pi2\right)}{-x^2} \, dx \\ &\quad - i \int_\infty^1 \frac{\left(\log\left\lvert1-x^2\right\rvert+i\pi\right)\left(\log x+i\frac{3\pi}2\right)}{-x^2} \, dx - i \int_1^\infty \frac{\left(\log\left\lvert1-x^2\right\rvert-i\pi\right) \left(\log x+i\frac{3\pi}2\right)}{-x^2} \, dx \\ &= -i2\pi \int_0^\infty \frac{\log\left(1+x^2\right)}{x^2} \, dx + i2\pi^2 \int_1^\infty \frac{dx}{x^2} \end{align*}$$

$$\implies \int_0^\infty \frac{\log\left(1+x^2\right)}{x^2} = \int_1^\infty \frac\pi{x^2} \, dx = \boxed\pi$$

It may be worth pointing out that the real technique (parts) works even more generally, as we have the exact anti-derivative

\begin{equation} \int \frac{\log(1+x^{2})}{x^{2}}dx = 2\arctan(x)-\frac{\log(x^{2}+1)}{x} \end{equation}

which can be verified by differentiating both sides with respect to x. This implies \begin{equation} \int_{-T}^{T}\frac{\log(1+x^{2})}{x^{2}}dx = 4\arctan(T)-2\frac{\log(T^{2}+1)}{T} \end{equation}

Then if you send $T \to \infty$ using $\arctan(T) \to \frac{\pi}{2}$, you get the claimed answer. Can the complex analysis techniques also be used to prove the finite T result here?

\begin{align} &\int_{0}^{\infty}{\ln(1 + x^{2}) \over x^{2}}\,\mathrm dx =\frac12\int_{0}^{\infty}{\ln(1 + x) \over x^{3/2}}\,\mathrm d x\\ &=-\frac12\int_{0}^{\infty}x^{-\frac32}\sum_{n=1}^\infty\frac1n (-x)^{n}\,\mathrm d x\\ &=\frac12\int_{0}^{\infty}x^{\frac12-1}\sum_{n=0}^\infty\frac1{n+1} (-x)^{n}\,\mathrm d x\\ &=\frac12\int_{0}^{\infty}x^{\frac12-1}\sum_{n=0}^\infty\frac{\Gamma(n+1)}{n+1} \frac{(-x)^{n}}{n!}\,\mathrm d x\\ &=\frac12\Gamma(\frac12)\frac{\Gamma(n+1)}{n+1}\bigg|_{n=-\frac12}\\ &=\pi. \end{align} By Ramanuja's Master Theorem.