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$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$
$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$
$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}-\int\limits_{0}^{1}\frac{dx}{1+x^2}$
Đặt $x=tanu$
$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$
$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$
$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du$
$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}
{4}}_{0}$
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