Compute integral $\int_1^3\frac{\ln x}{x^2+3}\ dx $ 
How to solve the integral $$\int_1^3\dfrac{\ln x}{x^2+3}\ dx\ ?$$ 

I was thinking of substituting $t=\ln x$ and then factor one $x$ out,is it correct?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}
A: 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: 
$$I=\color{red}{\frac{\pi\log 3}{12\sqrt{3}}}.\tag{3}$$
