Use inverse symmetry to integrate only half the lens

The curve increases from the horizontal asymptote $y=0$ as $x\to-\infty$ to the included endpoint $(5/2,4)$, and passes through $(0,2/13)$ and $(2,2)$. It is one-to-one by the horizontal-line test. For example $f^{\prime}(x)=-16(2x-5)/[(2x-5)^2+1]^2>0$ for $x<5/2$.

From $(2x-5)^2=4/y-1$ and $x\le5/2$, choose $2x-5=-\sqrt{4/y-1}$. Swapping variables gives $f^{-1}(x)=5/2-\tfrac12\sqrt{4/x-1}$ with domain $(0,4]$. The plus branch would invert a different restriction of the original rational curve.

The two curves reflect in $y=x$; between the intersections, $f(x)$ lies below the diagonal. Their enclosed area is $2\int_a^2[x-f(x)]dx=2[x^2/2-2\tan^{-1}(2x-5)]_a^2=4+\pi-a^2+4\tan^{-1}(2a-5)$ square units. The required result remains in terms of $a$.