How to get the following integral $ \int_0^s r^{-1/2}(s+t-2r)^{-3/2}dr $? How to estimate the following integral
$$
\int_0^s r^{-1/2}(s+t-2r)^{-3/2}dr
$$
where $t>>s>>1$.
It seems that for $t/s$ bounded, the above integral is approximate equal to $(t-s)^{1/2}$, otherwise $s^{3/4}/t^{5/4}$ for $t/s$ unbounded.

Following the comments, let $r=\frac{1}{2}(s+t)x$, then the integral is
$$
\frac{1}{\sqrt{2}}(s+t)^{-1}\int_0^{2s/(s+t)}  x^{-1/2}(1-x)^{-3/2}dx
$$
But how to use Beta function to get this integral in terms of $s$ and $t$?
 A: 
For $0<s<t$, we have
$$\begin{align}
I
&=\int_{0}^{s}\mathrm{d}r\,r^{-1/2}\left(s+t-2r\right)^{-3/2}\\
&=\frac{1}{\left(s+t\right)\sqrt{2}}\int_{0}^{\frac{2s}{s+t}}\mathrm{d}x\,x^{-1/2}\left(1-x\right)^{-3/2};~~~\small{\left[r=\frac12(s+t)x\right]}\\
&=\frac{1}{\left(s+t\right)\sqrt{2}}\int_{0}^{z}\mathrm{d}x\,\frac{1}{\sqrt{x\left(1-x\right)^{3}}};~~~\small{\left[z:=\frac{2s}{s+t}\in(0,1)\right]}\\
&=\frac{1}{\left(s+t\right)\sqrt{2}}\int_{0}^{z}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}\sqrt{\frac{x}{1-x}}}\\
&=\frac{1}{\left(s+t\right)\sqrt{2}}\int_{0}^{\frac{z}{1-z}}\mathrm{d}y\,\frac{1}{\sqrt{y}};~~~\small{\left[\frac{x}{1-x}=y\implies x=\frac{y}{1+y}\right]}\\
&=\frac{\sqrt{2}}{s+t}\sqrt{\frac{z}{1-z}}\\
&=\frac{\sqrt{2}}{s+t}\sqrt{\frac{2s}{t-s}}\\
&=\frac{2}{s+t}\sqrt{\frac{s}{t-s}}.\blacksquare\\
\end{align}$$

A: Letting $r=\frac{s+t}{2}x$ yields
$$
I=\frac{1}{\sqrt{2}\left(s+t\right)} \int_0^{\frac{2 s}{s+t}} x^{-\frac{1}{2}}(1-x)^{-\frac{3}{2}} d x
$$
For getting rid of surds, we substitute $x=\sin^2\theta$ and obtain $$
\int x^{-\frac{1}{2}}(1-x)^{-\frac{3}{2}} d x=2 \int \sec ^2 \theta d \theta=2 \tan \theta+C=2 \sqrt{\frac{x}{1-x}} +C
$$
$$
I=\frac{1}{\sqrt{2}(s+t)}\left[2 \sqrt{\frac{x}{1-x}}\right]_0^{\frac{2 s}{s+t}}=\frac{2}{s+t} \sqrt{\frac{s}{t-s}}
$$
A: As pointed out in the comments, the beta function is defined as follows:
$$
B(x;\alpha,\beta)=\int_0^{x}  t^{\alpha-1}(1-t)^{\beta-1}dt
$$
Your integral:
$$
\int_0^{\dfrac{2s}{s+t}}  x^{1/2-1}(1-x)^{-1/2-1}dx=B(\dfrac{2s}{s+t};\dfrac{1}{2},\dfrac{-1}{2})
$$
