Question

$\underset{0}{\overset{1}{\int }}\frac{1-x^2}{{\left(1+x^2\right)}^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{1-x^2}{{\left(1+x^2\right)}^2}\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_0^1\frac{\left({\displaystyle \frac{1}{x^2}}-1\right)}{{\left(x+{\displaystyle \frac{1}{x}}\right)}^2}dx \\ \mathrm{Let}x+\frac{1}{x}=t.\mathrm{Then},1-\frac{1}{x^2}dx=dt \\ \mathrm{When}x=0,t=\infty \mathrm{and}x=1,t=2 \\ \therefore I={\int }_{\infty }^2\frac{-dt}{t^2} \\ \Rightarrow I={\left[\frac{1}{t}\right]}_{\infty }^2 \\ \Rightarrow I=\frac{1}{2}-0 \\ \Rightarrow I=\frac{1}{2} \end{gathered}$$