$\begin{aligned} & \mathrm{IF}=\mathrm{e}^{\int \frac{\mathrm{zx}}{\left(1+\mathrm{x}^2\right)^2} \mathrm{dx}}=\mathrm{e}^{\frac{-1}{1+\mathrm{x}^2}} \\ & \mathrm{y} \cdot \mathrm{e}^{\frac{-1}{1+\mathrm{x}^2}}=\int \mathrm{x} \cdot \mathrm{e}^{\frac{1}{1+\mathrm{x}^2}} \cdot \mathrm{e}^{\frac{-1}{1+\mathrm{x}^2} \mathrm{dx}}\end{aligned}$
$\begin{aligned} & y \cdot e^{\frac{-1}{1+x^2}}=\frac{x^2}{2}+c \\ & (0,0) \Rightarrow C=0 \\ & y(x)=\frac{x^2}{2} e^{\frac{1}{1+x^2}} \\ & f(x)=\frac{x^2}{2}\end{aligned}$

$A=\int_{-2}^4(x+4)-\frac{x^2}{2} d x=18$