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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}+3\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}$

$\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}$

$\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}+3\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$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du+3\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du+4\int\limits_{0}^{\frac{\pi }{4}}du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}+4u|^{\frac{\pi}{4}}_{0} $

$\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}+3\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}$

$\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$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0} $

$\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}+3\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$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du+3\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du+4\int\limits_{0}^{\frac{\pi }{4}}du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}+4u|^{\frac{\pi}{4}}_{0} $