I am trying to show that for $0<a<b$,
$$\int_a^b x \arctan \sqrt{ \left( \frac{a+b}{\frac{ab}{x}+x} \right)^2 - 1} \text{d}x = \frac{\pi (b-a)^2}{8}.$$
I have reason to believe this is true and numerically it checks out for the examples I have tried. This integral appears to have a geometric interpretation. But I'm unable to solve it directly and Wolfram Alpha cannot handle it. What is a good approach for this integral?
Although not very conspicuous, the integral is elementary
$$\begin{aligned} \int {x\arctan \sqrt {{{\left( {\frac{{(a + b)x}}{{ab + {x^2}}}} \right)}^2} - 1} dx} &= \text{ product from by parts } + \frac{1}{2}\int {{x^2}\frac{{ab - {x^2}}}{{(abx + {x^3})\sqrt {{{\left( {\frac{{(a + b)x}}{{ab + {x^2}}}} \right)}^2} - 1} }}dx} \\ &= \cdots+\frac{1}{2}\int {x\frac{{ab - {x^2}}}{{\sqrt {{{(a + b)}^2}{x^2} - {{(ab + {x^2})}^2}} }}dx} \\ &= \cdots +\frac{1}{4}\int {\frac{{ab - u}}{{\sqrt {{{(a + b)}^2}u - {{(ab + u)}^2}} }}du} \end{aligned}$$ where $u = x^2$.
The result of definite integral is particular elegant, partly, because the $\cdots$ terms vanishes when the limits of integration are well-chosen.
