Area bounded by the curves  $\mathrm{y}=x^2$ and  $\mathrm{y}=\frac{2}{1+x^2}$ is

$$\begin{gathered} \mathrm{y}=x^2 and y=\frac{2}{1+x^2}\Rightarrow y=\frac{2}{1+y} \\ \Rightarrow \mathrm{y}+y^2=2 \\ \Rightarrow (y-1)(y+2)=0 \\ \mathrm{Y}=-2,1 but y\ge 0 so \mathrm{y}=1\Rightarrow x=\pm 1 \end{gathered}$$

Since area is symmetric about y-axis 
$\therefore$ Required area  $=2[\underset{0}{\overset{1}{\int }}\frac{2}{1+x^2}dx-\underset{0}{\overset{1}{\int }}{x}^{2}dx]$
$=2\{{\left(2{\tan }^{-1}x\right)}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\}=2[\frac{2.\pi }{4}-\frac{1}{3}]=\pi -\frac{2}{3}$