Compute integral $\int_1^3\frac{\ln x}{x^2+3}\ dx $

By setting $x=\frac{3}{y}$ we get $$ I = \int_{1}^{3}\frac{\log x}{x^2+3}\,dx = \int_{1}^{3}\frac{\log(3)-\log(y)}{y^2+3}\,dy\tag{1}$$ hence it follows that: $$ 2I = \int_{1}^{3}\frac{\log 3}{z^2+3}\,dz = \frac{\pi\log 3}{6\sqrt{3}}\tag{2} $$ and symmetry wins again: $$I=\color{red}{\frac{\pi\log 3}{12\sqrt{3}}}.\tag{3}$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{1}^{3}{\ln\pars{x} \over x^{2} + 3}\,\dd x & \,\,\,\stackrel{x/\root{3}\ \mapsto\ x}{=}\,\,\, {\root{3} \over 3} \int_{\root{3}/3}^{\root{3}}{\ln\pars{\root{3}x} \over x^{2} + 1}\,\dd x \\[1cm] & = {\root{3} \over 3}\,\ln\pars{\root{3}} \int_{\root{3}/3}^{\root{3}}{\dd x \over x^{2} + 1} \\[3mm] & + {\root{3} \over 6}\bracks{% \int_{\root{3}/3}^{\root{3}}{\ln\pars{x} \over x^{2} + 1}\,\dd x + \int_{\root{3}/3}^{\root{3}}{\ln\pars{x} \over x^{2} + 1}\,\dd x} \\[1cm] & = {\root{3}\ln\pars{3} \over 6}\bracks{\arctan\pars{\root{3}} - \arctan\pars{\root{3} \over 3}} \\[3mm] & + {\root{3} \over 6}\ \underbrace{\bracks{% \int_{\root{3}/3}^{\root{3}}{\ln\pars{x} \over x^{2} + 1}\,\dd x + \int_{3/\root{3}}^{1/\root{3}}{\ln\pars{1/x} \over 1/x^{2} + 1} \pars{-\,{1 \over x^{2}}}\,\dd x}}_{\ds{=\ 0}} \\[1cm] & = {\root{3}\ln\pars{3} \over 6}\pars{{\pi \over 3} - {\pi \over 6}} = \bbx{{\root{3}\ln\pars{3} \over 36}\,\pi} \end{align}