Let $y=f\left(x\right)$  defined on R satisfies the differential equation  $(1+x^2)\frac{dy}{dx}+2xy=2x$and $f\left(0\right)=2,$ then which of the following statement(s) is (are) correct?

$\begin{array}{r}\left(1+x^2\right)\frac{dy}{dx}+2xy=2x\\ \Rightarrow \frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{2x}{1+x^2}\end{array}$

$\text{ which is a linear differential equation. }$

$\begin{array}{r}\int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}=\ln \left(1+{\mathrm{x}}^2\right)\\ \therefore \text{ l. }F.={\mathrm{e}}^{\ln \left(1+{\mathrm{x}}^2\right)}=1+{\mathrm{x}}^2\end{array}$

$\text{ Now, the general solution is }$

$\begin{array}{l}y\left(1+x^2\right)=\int \left(\frac{2x}{1+x^2}\right)\cdot \left(1+x^2\right)dx+C\\ \Rightarrow y\left(1+x^2\right)=x^2+C\\ y\left(0\right)=2\Rightarrow 2=0+C\\ \Rightarrow C=2\\ \therefore f\left(x\right)=\frac{x^2+2}{x^2+1}\end{array}$

$f\left(x\right)=1+\frac{1}{x^2+1}$

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=\frac{-2x}{{\left(x^2+1\right)}^2}\\ \therefore f^{\mathrm{\prime }}\left(x\right)>0\mathrm{\forall }x<0\text{ and }{f}^{\mathrm{\prime }}\left(x\right)<0\mathrm{\forall }x>0\end{array}$

$\text{ Hence, }f\left(x\right)\text{ increases on }(-\mathrm{\infty },0)\text{ and decreases on }(0,\mathrm{\infty })\text{. }$

$\begin{array}{r}\text{ Area }={\int }_0^1 f\left(x\right)dx\\ ={\int }_0^1 \left(1+\frac{1}{1+x^2}\right)dx\\ ={\left[x+{\tan }^{-1}x\right]}_0^1\\ =\left(\frac{\pi +4}{4}\right)\text{ sq. units }\end{array}$