The following reslut is from a paper,The author says it is not hard to show that:

$$\lim_{t\to 1}\dfrac{1-t}{\sqrt{1+pt}}\int_{0}^{t}\dfrac{a(1+pa)}{(1-a)^2}\left(4a\left[1-\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}\right]\right)^{-1/2}da=\dfrac{\pi}{4}\sqrt{p+1}$$

But I try use Taylor formula $$\left(4a\left[1-\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}\right]\right)^{-1/2}=\dfrac{1}{2\sqrt{a}}\left(1-\dfrac{1}{2}\left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt}+o( \left(\dfrac{1-t}{1-a}\right)^2\dfrac{1+pa}{1+pt})\right)$$ still can't Solve this problem ,so How to solve this problem Thanks