Theorem (*): 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 $$\int f ^{-1} (x) dx= xf ^{-1} (x)+ F(f ^{-1}(x))+C$$
Example: Calculate $ \displaystyle \int \dfrac{1-\sqrt{1-x^2}}{x}dx$ where $x\in[-1,1]$
Solution. Consider $f(x)=\dfrac{2x}{x^2+1}$ be differentiable ,monotonic function on $[-1,1]$ and $ \displaystyle \int \dfrac{2x}{x^2+1}dx=\ln (x^2+1)+C$
We have $f^{-1}(x)= \dfrac{1-\sqrt{1-x^2}}{x}$, by (*) we get
$$ \displaystyle \int \dfrac{1-\sqrt{1-x^2}}{x}dx$$ $$ =x\cdot \left(\dfrac{1-\sqrt{1-x^2}}{x}\right)-\ln \left(\left(\dfrac{1-\sqrt{1-x^2}}{x}\right)^2+1 \right)+C$$
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 $\int g(x) dx$
An other similar question posted in Integral of reflected function
Thank you in advance!
