Use inverse symmetry to integrate only half the lens

Define $f(x)=4/[(2x-5)^2+1]$ for real $x\le5/2$.

(i) Sketch $y=f(x)$ and show $f^{-1}$ exists.

(ii) Find $f^{-1}(x)$ and its domain.

The intersections of $f$ and $f^{-1}$ have x-coordinates $a$ and $2$, where $a<2$. Find the enclosed area in terms of $a$.