What is the fallacy when trying to evaluate the FT of the dirac delta function via integration by parts?

Question

When evaluating the Fourier transform of the dirac delta function, if I call the dirac delta function dv as a basis to try to integrate by parts, then because the integral of the dirac delta function is 1 = v then all that is left is u - the integral of du which would reduce to 0. But the actual FT is 1. The same question would apply to ANY function with an integral of 1 across the real line (i.e. finite signal with total AOC of 1); there's no way the FTs are just 0. Where is the mistake in this reasoning? Is there an approach to evaluate the FT for any given function with an integral of 1?