Can a change of variable result in the evaluation of an integral in terms of elementary functions, whereas before the c.o.v. this was not possible?

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Suppose we have an integral $$I=\int f(x)dx,$$ for some function $f$ which can not be evaluated in terms of elementary functions, e.g. in terms of exponentials, a finite number of logs, and algebraic functions (see Liouville's theorem). An example of such a function is $f(x)=e^{-x^2}$.

I am wondering if there exists such a function $f$ for which, after some change of variable $x=g(y)$, the integral $$J=\int f(g(y))g'(y)dy,$$ can be evaluated in terms of elementary functions? Is there an example of such a case? Or maybe such integrals are "closed" within their own space of non-elementary functions.

Clarification

I wish to know whether or not, for a function $f $ whose primitive is non elementary, a substitution could result in a function $f\circ g $ whose primitive is elementary. The two different integrands will produce different graphs so they are not the same function. Yes their area under the curve will be the same. But I'm wondering if the change of variable would make the substituted integrand "amenable" to antidifferentiation in terms of elementary functions. I'm thinking no because any transformed function will still be in the same differential field, although not sure.

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Trivially, yes... with a non-elementary change of variable. But undo the change of variable and you will see that the primitive remains non-elementary.

Let be $F(x) = \int f(x)\,dx$. With the cov $x = F^{-1}(y)$, $$\int f(x)\,dx = \int f(F^{-1}(y))(F^{-1})'(y)\,dy = \int F'(F^{-1}(y))(F^{-1})'(y)\,dy = \int 1\,dy =y.$$ (by the IFT $(F^{-1})'(y) = 1/F'(F^{-1}(y))$)