Prove that if $\int f(x) dx$ is non elementary then $\int f^{-1}(x) dx$ is also non elementary I asked this question yesterday but I can't understand a few things.
Proof of this theorem about non-elementary integrals
Here I'm assuming $F(x) = \int f(x)\:dx$ is non elementary.
If you make $u = f^{-1}(x)$ then $du = (f^{-1}(x))' \:dx$. I understantd that $f'(u)=1$.
So, $\int (f^{-1}(x))\:dx $ = $\int \frac{u}{u'}f'(u)\:du $. If $u'$ is the derivative of $u$ with respect to $x$.
How I get to this result $\int u f'(u) \:du$ ? And $F(u) = \int f(u)\:du$?
If somenone can make a way more detailed proof I would appreciate.  
 A: I tried to reformulate the original answer with more detail. Be careful : we are assuming that $F(x)=\int f(x) \mathrm{d}x$ is elementary. Our goal is to prove that $G(x)=\int f^{-1}(x)\mathrm{d}x$ is also elementary.
Using a u-substitution $u=f^{-1}(x)$, we have $x=f(u)$ so $\mathrm{d}x=f'(u)\mathrm{d}u$ and
$$
G(x)=\int u f'(u) \mathrm{d}u.
$$
Alternatively, you can differentiate the expression $u=f^{-1}(x)$ (this is slightly more complicated). You get $\mathrm{d}u=\dfrac{1}{f'(f^{-1}(x))}\mathrm{d}x=\dfrac{1}{f'(u)}\mathrm{d}x$, which is obviously the same result.
Now we integrate by parts :
$$
\begin{align}
G(x)&=\int uf'(u)\,\mathrm{d}u \\
&= uf(u)-\int f(u) \,\mathrm{d}u \\
&=uf(u)-F(u) \\
&=xf^{-1}(x)-F(f^{-1}(x)),
\end{align}
$$
so $G$ is elementary.
A: *

*Suppose $f$ and $g$ are inverses of each other. 

*Because they're inverses, $(f\circ g) = \text{identity}$. Hence by the chain rule, $(f\circ g)^\prime = 1 = (f^\prime \circ g)\cdot g^\prime$, so $(f^\prime \circ g) = \frac{1}{g^\prime}$.

*Let $F(x)=\int_0^x f(t)dt$ and $G(x)=\int_0^x g(t)dt $ be antiderivatives of $f$ and $g$, respectively.

*Using integration by substitution ("u-substitution"), we know that 
$$G(x) = \int_0^x g(t) dt = \int_{f^{-1}(0)}^{f^{-1}(x)} (g\circ f)(u) \cdot f^{\prime}(u) du$$
but $f^{-1} = g$ and also $g\circ f= \text{identity}$, because $g$ and $f$ are inverses. Hence:
$$G(x) = \int_{g(0)}^{g(x)} u\cdot f^\prime(u)\,du$$


*To evaluate this integral, let $h(u)=u$ be the identity function and consider the product rule: $(f\cdot h)^\prime = f^\prime \cdot h + f\cdot h^\prime$. If we integrate both sides from $a$ to $b$, we find:


$$[f\cdot h]_{a}^b = \int_{a}^b f^\prime\cdot h + \int_{a}^b f\cdot h^\prime$$


*But of course $h^\prime = 1$. Hence we find a simplification:
$$[f\cdot h]_{a}^b = \int_{a}^b f^\prime\cdot h + \int_{a}^b f$$
or 
$$\int_a^b f^\prime \cdot h = [f\cdot h]_a^b - [F(b)-F(a)]$$

*Hence 
$$G(x) = \int_{g(0)}^{g(x)} u\cdot f^\prime(u)\,du = [f\cdot h]_{g(0)}^{g(x)} - [F(g(x)) - F(g(0))]$$
$$G(x) = g(x)\cdot f(g(x)) - g(0) \cdot f(g(0)) - F(g(x)) + F(g(0))$$ 
$$G(x) = x g(x) - 0 - F(g(x)) + F(g(0))$$
$$G(x) = xg(x) - F(g(x)) \;\;\;+\underbrace{F(g(0))}_{=const.}$$

*Now we bring in our elementariness assumption: suppose $f$ and $g$ are elementary. If $F$ is elementary, then so are $x\cdot g(x)$, $(F\circ g)$, and so on. Hence $G$ is elementary.

*We could have done this proof with the names $f$ and $g$ exchanged  (along with $F$ and $G$) . By switching the role of $f$ and $g$ in our proof, we get an exactly analogous result: if $G$ is elementary, then so is $F$.

*Hence, assuming $f$ and $g$ are elementary, $F$ is elementary if and only if $G$ is.
