^ Even if it does, none of that helps you integrate the original function. As I said before, I'm fairly sure you can't just go around making changes to the function that aren't equal to 1, and expect the same answer. (Its why when you're doing some integrals, you take natural logs AND exponentials at the same time, because log(exp f(x))= f(x), or when you're working with complex numbers, you multiply by the complex conjugate/ complex conjugate, because its equal to 1.) if you take a natural log, without taking an exponential, you've changed the function, so the integral won't give you the correct answer for the function anymore.
E.g. If I just take the logarithm of some function, for simplicity, lets say y=2x
y=2x
ln(y)=ln(2x)
int(ln(y))=int(ln(2x))
yln(y) - y = ln(2x)*x -x +C
Which clearly is not equal to int (y) = int(2x) [ y^2/2 = x^2+C], because, for one thing, the natural logarithms are undefined for x or y less than 0, where clearly the original function doesn't have that problem
E.g. If I just take the logarithm of some function, for simplicity, lets say y=2x
y=2x
ln(y)=ln(2x)
int(ln(y))=int(ln(2x))
yln(y) - y = ln(2x)*x -x +C
Which clearly is not equal to int (y) = int(2x) [ y^2/2 = x^2+C], because, for one thing, the natural logarithms are undefined for x or y less than 0, where clearly the original function doesn't have that problem