Functional Integral

$\displaystyle{xf(x) = \int{f(x)(x+1)dx} \implies \dfrac{d}{dx}\left[xf(x)\right] = \dfrac{d}{dx}\left[\displaystyle{\int{xf(x)+f(x)dx}}\right] \\ \dfrac{d}{dx}\left[xf(x)\right] = xf(x)+f(x) \implies f(x)+xf'(x) = xf(x)+f(x) \\ \therefore f(x) = f'(x) \implies f(x) = \space ae^{x}}$

We have that $f(0) = 1$ and therefore $a = 1$ so $e^0$ is indeed $1$ .

$\text{Thus},\quad f(2) = e^2 \approx 7.389$

The constant of integration can be nothing but $0$ here because $f(x) = f'(x)$ and this wouldn't hold if $f(x) = e^x + C$ for $C \neq 0$ .