Calculate the inverse for $\arctan(x^2+1), x≥0$

Question

I have no idea how to solve this problem.

Calculate the inverse for the function:

$$f(x) = \arctan(x^2+1),\quad x≥0 .$$

Also specify $D_{f^{-1}}$ and $V_{f^{-1}}$.

I would really appreciate your help

Answer 1

$$f(x)=y = \arctan(x^2+1), x≥0$$

$$\tan y =\tan( \arctan(x^2+1))=x^2+1$$

$$x^2=\tan y -1$$

$$x=\sqrt{\tan y-1}, \tan y\geq 1$$