I want to show that $$\frac{1-t}{2\pi} \int\_{a}^{b} \frac{\sqrt{(b-x)(x-a)}}{s+tx}dx = \frac{1}{1-t},$$ where $a = \frac{(1-h)^2}{(1-t)^2}, b = \frac{(1+h)^2}{(1-t)^2}$ and $h = \sqrt{s+t-st}$. 

I have tried the following so far: Let us manipulate the integrand into something reasonable to deal with. To this end, we have \begin{eqnarray\*}
a+b &=& \frac{(1-h)^2}{(1-t)^2} + \frac{(1+h)^2}{(1-t)^2} = \frac{1-2h+h^2 + 1 + 2h + h^2}{(1-t)^2} = 2\frac{h^2+1}{(1-t)^2}.
\end{eqnarray\*} Similarly, we have \begin{eqnarray\*}
ab &=& \frac{(1-h)^2(1+h)^2}{(1-t)^4} = \frac{(1-h^2)^2}{(1-t)^4}.
\end{eqnarray\*} Thus, \begin{eqnarray\*}
\sqrt{(b-x)(x-a)} &=& \sqrt{(a+b)x - ab - x^2} \\
&=& \sqrt{\frac{2x(h^2+1)}{(1-t)^2} - \frac{(1-h^2)^2}{(1-t)^4} - x^2} \\
&=&\sqrt{- \left( \left( x - \frac{(h^2+1)}{(1-t)^2} \right)^2 - \frac{(h^2+1)^2 + (1-h^2)^2}{(1-t)^4} \right)} \\
&=& \sqrt{\frac{2h^4 + 2}{(1-t)^4} - \left( x - \frac{(h^4+1)}{(1-t)^2} \right)^2} \\
&=& \sqrt{\frac{2h^4 + 2}{(1-t)^4} - \left( \frac{(1-t)^2 x - (h^4+1)}{(1-t)^2} \right)^2} \\
&=& \frac{1}{(1-t)^2} \sqrt{2h^4 + 2 - [(1-t)^2 x - (h^4+1) ]^2 }. 
\end{eqnarray\*} So we currently have \begin{eqnarray\*}
\int \frac{\sqrt{(b-x)(x-a)}}{s+tx}dx &=& \frac{1}{(1-t)^2} \int \frac{\sqrt{2h^4 + 2 - [(1-t)^2 x - (h^4 + 1)]^2}}{s+tx}dx 
\end{eqnarray\*} Now let us attempt to make the substitution $$u = 2h^4 + 2 - [(1-t)^2 x - (h^4 + 1)]^2.$$ From this, we see that \begin{eqnarray\*}
du = -2(1-t)^2[(1-t)^2 x - (h^4+1)] dx,
\end{eqnarray\*} and \begin{eqnarray\*}
x &=& \frac{-(h^4+1) \pm \sqrt{2(h^4 + 1) - u}}{2(1-t)^2}. 
\end{eqnarray\*} Inserting this into our expression for the integral, we have \begin{eqnarray\*}
&&\frac{1}{(1-t)^2} \int \frac{\sqrt{2h^4 + 2 - [(1-t)^2 x - (h^4 + 1)]^2}}{s+tx}dx \\
&=& \frac{1}{(1-t)^2} \int \frac{\sqrt{u}}{s+t \left( \frac{-(h^4+1) \pm \sqrt{2(h^4 + 1) - u}}{2(1-t)^2} \right) } \frac{du}{-2(1-t)^2[(1-t)^2 \left( \frac{-(h^4+1) \pm \sqrt{2(h^4 + 1) - u}}{2(1-t)^2} \right) - (h^4+1)]} \\
&=& \frac{1}{(1-t)^2} \int \frac{\sqrt{u}}{s+t \left( (h^4+1) \pm \sqrt{2(h^4+1)-u} \right)[(1-t)^2 \left( \frac{-(h^4+1) \pm \sqrt{2(h^4 + 1) - u}}{2(1-t)^2} \right) - (h^4+1)]}du \\
&=& \frac{1}{(1-t)^2} \int \frac{\sqrt{u}}{s+ t \left( (h^4+1) \pm \sqrt{2(h^4+1) -u} \right) [- \frac{1}{2}(h^4+1) \pm \frac{1}{2} \sqrt{2(h^4+1) -u} - (h^4+1) ] }
\end{eqnarray\*}

There must be a quicker way to solve this...