Integral of a function's derivative does not equal the original function?

Assuming $f$ is differentiable, then the fundamental theorem of calculus says

$$\int_3^xf'(s) \, ds=f(x)-f(3)$$

Hence, unless $f(3)=0$, the integral expression is not $f(x)$.

Actually, I disagree with your statement 'the left side yields...'

You are talking about indefinite integrals, but here you have a definite integral. In particular, you have $$ \int_3^x f'(x)\,dx=g(x)-g(3), $$ where $g$ is any antiderivative of $f'(x)$. In particular, we know that all antiderivatives of $f'(x)$ are of the form $f(x)+C$ for some constant $C$, so that $$ \int_3^x f'(x)\,dx=[f(x)+C]-[f(3)+C]=f(x)-f(3). $$ So, your question boils down to this: is $f(x)=f(x)-f(3)$ true for all $x$? The answer will depend on the particular value that your function $f$ assigns to the input $3$.