Let $f(x)$ be differentiable , monotonic function, $f^{-1}(x)$ is inverse function of $f(x)$ and knowning that $\int f(x) dx= F(x)+C$. Prove that $$\int f ^{-1} (x) dx= xf ^{-1} (x)+ F(f ^{-1}(x))+C$$
Proof. Note that $f( f ^{-1}(x))=x$ so $[ f( f ^{-1}(x))]'=1\Leftrightarrow f'(f ^{-1}(x)) (f ^{-1}(x))' =1\Rightarrow (f ^{-1}(x))'= \dfrac{1}{ f'(f ^{-1}(x))}$
So $$ \left(xf ^{-1} (x)+ F(f ^{-1}(x))\right)'= f ^{-1} (x)+ \dfrac{x}{ f'(f ^{-1}(x))} - \dfrac{ f(f ^{-1}(x)) }{ f'(f ^{-1}(x))}$$ $$ = f ^{-1}(x)+ \dfrac{x}{ f'(f ^{-1}(x))} - \dfrac{x}{ f'(f ^{-1}(x))} = f ^{-1}(x). $$ So $ \displaystyle \int f ^{-1} (x) dx= xf ^{-1} (x)+ F(f ^{-1}(x))+C$.
Question: I just think how we can solve the following generaztion: Let $f(x)$ be differentiable , monotony function, $g(x)$ is reflected function of $f(x)$ through line $ax+by+c=0$ and knowning that $\int f(x) dx= F(x)+C$. Computer $\int g(x) dx$
My efforts: I think we should use the reflection formula for above generalization : $\dfrac{x'-x}a=\dfrac{y'-y}b=\dfrac{-2(ax+by+c)}{a^2+b^2}$ in How to find the equation of the graph reflected about a line?
Thank you in advance!
