1. Solve the differential equation

Given $\frac{dy}{dx}+\frac{2x}{(1+x^2)^2}y=x e^{\frac{1}{1+x^2}}, \qquad y(0)=0.$

This is a linear differential equation: $\frac{dy}{dx}+P(x)y=Q(x),$ with $P(x)=\frac{2x}{(1+x^2)^2}, \qquad Q(x)=x e^{\frac{1}{1+x^2}}.$

2. Find the integrating factor

We compute $\int P(x)\,dx=\int \frac{2x}{(1+x^2)^2}\,dx.$ Let $u=1+x^2$, so $du=2x\,dx$. Then $\int \frac{2x}{(1+x^2)^2}\,dx=\int u^{-2}\,du=-u^{-1}=-\frac{1}{1+x^2}.$

Hence the integrating factor is $\mathrm{IF}=e^{\int P(x)dx}=e^{-\frac{1}{1+x^2}}.$

3. Multiply the equation by the integrating factor

Then $e^{-\frac{1}{1+x^2}}\frac{dy}{dx}+\frac{2x}{(1+x^2)^2}e^{-\frac{1}{1+x^2}}y=x.$

So the left side becomes $\frac{d}{dx}\left(y e^{-\frac{1}{1+x^2}}\right)=x.$

But the problem defines $f(x)=y(x)e^{-\frac{1}{1+x^2}}.$ Thus, $f'(x)=x.$

4. Integrate and use the initial condition

Integrating, $f(x)=\frac{x^2}{2}+C.$

Now use $y(0)=0$: $f(0)=y(0)e^{-1}=0 \implies C=0.$ Thus $f(x)=\frac{x^2}{2}.$

So the curve is $y=\frac{x^2}{2}.$

5. Find intersection with the line

The given line is $y-x=4 \quad \Rightarrow \quad y=x+4.$

Intersection points satisfy $\frac{x^2}{2}=x+4.$ Multiply by $2$: $x^2=2x+8,$ $x^2-2x-8=0,$ $(x-4)(x+2)=0.$ So the points of intersection are at $x=-2,\;4.$

6. Determine which curve is above

Consider $\left(x+4\right)-\frac{x^2}{2}.$ At $x=0$, this is $4>0$, so the line lies above the parabola on $[-2,4]$.

Hence enclosed area is $A=\int_{-2}^{4}\left[(x+4)-\frac{x^2}{2}\right]dx.$

7. Evaluate the area

$A=\int_{-2}^{4}\left(x+4-\frac{x^2}{2}\right)dx.$

Antiderivative: $\int \left(x+4-\frac{x^2}{2}\right)dx=\frac{x^2}{2}+4x-\frac{x^3}{6}.$

Now evaluate from $-2$ to $4$:

At $x=4$, $\frac{4^2}{2}+4(4)-\frac{4^3}{6}=8+16-\frac{64}{6}=24-\frac{32}{3}=\frac{40}{3}.$

At $x=-2$, $\frac{(-2)^2}{2}+4(-2)-\frac{(-2)^3}{6}=2-8+\frac{8}{6}=-6+\frac{4}{3}=-\frac{14}{3}.$

Therefore, $A=\frac{40}{3}-\left(-\frac{14}{3}\right)=\frac{54}{3}=18.$

8. Final answer

The enclosed area is $\boxed{18}.$