Lovely problem

Evaluating the given equation at $x=0$ , we find that $f(0)=0$ .

The second fundamental theorem gives $f'(x)=f(x)$ , or $f'(x)-f(x)=0$ . Multiplying with the integrating factor $e^{-x}$ , we can conclude that $0=f'(x)e^{-x}-f(x)e^{-x}=\frac{d}{dx}(f(x)e^{-x})=0$ . Thus $f(x)e^{-x}=k$ for some constant $k$ . Now $k=0$ since $f(0)=0$ , so that $f(x)=0$ for all $x$ . In particular, $f(\ln{5})=\boxed{0}$

$\large f(x) = \displaystyle\int_0^x f(t) \text{dt} \quad \quad ..\color{#3D99F6}{\boxed{1}}$ Applying Newton's Leibniz rule

$\Rightarrow f'(x) = f(x)$
$\Rightarrow \dfrac{f'(x)}{f(x)} = 1$
Integration both sides with respect to $\text{dx}$

$\Rightarrow \displaystyle\int \dfrac{f'(x)}{f(x)}\text{dx} = \displaystyle\int \text{dx}$
$\Rightarrow \ln| f(x) | = x + c$
$\Rightarrow f(x) = ke^x \quad \quad ..\color{#D61F06}{\boxed{2}}$
Substituting value of $\color{#D61F06}2$ in equation $\color{#3D99F6}1$ ,

$\Rightarrow e^x = \displaystyle\int_0^x ke^t \text{dt}$
Again applying Newton's leibniz rule,

$\Rightarrow ke^x = k(e^x - 1) \Rightarrow k = 0$
$\therefore f(x) = 0 \Rightarrow \boxed{f(\ln 5) = 0}$