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$\begin{array}{l}
f\left( x \right) = \dfrac{{2x}}{{{x^2} + 1}} \Rightarrow {f^2}\left( x \right) = \dfrac{{4{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}\\
\bullet \left( {1 - {x^2}} \right){f^2}\left( x \right).g\left( {f\left( x \right)} \right) = 4{x^2}g\left( x \right)\\
\Leftrightarrow \left( {1 - {x^2}} \right).\dfrac{{4{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}g\left( {f\left( x \right)} \right) = 4{x^2}g\left( x \right)\\
\Rightarrow \dfrac{{\left( {1 - {x^2}} \right)g\left( {f\left( x \right)} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} = g\left( x \right)\\
\Leftrightarrow \dfrac{{\left( {1 - {x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}g\left( {f\left( x \right)} \right) = g\left( x \right)\\
\Leftrightarrow \dfrac{{g\left( {f\left( x \right)} \right)}}{{g\left( x \right)}} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{\left( {1 - {x^2}} \right)}}
\end{array}$
Chọn hàm $g(x)=\dfrac{1}{1-x^2}$
Thử lại:
$\begin{array}{l}
g\left( {f\left( x \right)} \right) = \dfrac{1}{{1 - {{\left( {\dfrac{{2x}}{{{x^2} + 1}}} \right)}^2}}} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\\
\Rightarrow \dfrac{{g\left( {f\left( x \right)} \right)}}{{g\left( x \right)}} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{\left( {1 - {x^2}} \right)}}
\end{array}$
$\begin{array}{l}
\int\limits_0^{\dfrac{1}{2}} {xg\left( x \right)dx} = \int\limits_0^{\dfrac{1}{2}} {\dfrac{x}{{1 - {x^2}}}dx = - \dfrac{1}{2}\int\limits_0^{\dfrac{1}{2}} {\dfrac{{2x}}{{1 - {x^2}}}dx} } \\
= - \dfrac{1}{2}\int\limits_0^{\dfrac{1}{2}} {\dfrac{{d\left( {1 - {x^2}} \right)}}{{1 - {x^2}}} = \ln \dfrac{{2\sqrt 3 }}{3}} \\
\Rightarrow \int\limits_0^{\dfrac{1}{2}} {g\left( x \right)} = \int\limits_0^{\dfrac{1}{2}} {\dfrac{1}{{1 - {x^2}}}dx} = \dfrac{1}{2}\int\limits_0^{\dfrac{1}{2}} {\dfrac{2}{{1 - {x^2}}}dx} = \dfrac{1}{2}\int\limits_0^{\dfrac{1}{2}} {\left( {\dfrac{1}{{1 - x}} + \dfrac{1}{{1 + x}}} \right)} dx\\
= \left. {\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right)} \right|_0^{\dfrac{1}{2}} = \ln \sqrt 3
\end{array}$
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