<s>[INST] $\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$ 
Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
 1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
 {\rm d}t=\dfrac{\pi }{\sqrt{2}}$.

We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?
 [/INST] Finally I was able to handle this problem. It is just tedious calculation. Let $u=\frac{f(t)}{f(x)}$ and then
\begin{eqnarray}t&=&1-\frac{(1-x)(x-1+\sqrt{(1-x)^2+8u(1+x)}}{2u(1+x)},\\
dt&=&-\frac12\frac{ (x-1) \left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)}{u^2(1+x)\sqrt{(1-x)^2+8u(1+x)}}du. 
\end{eqnarray}
So
\begin{eqnarray}
&&\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t\\
&=&\lim_{x \to 1}\int_{\frac1{f(x)}}^1\frac{\sqrt{t}\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
   }{u \sqrt{(1-u)
   (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}{\rm d}u\\
&=&\int_0^1\frac{1}{\sqrt{2u(1-u)}}du\\
&=&\frac{\pi}{\sqrt 2}.
\end{eqnarray}
Here
$$ \lim_{x \to 1}\frac1{f(x)}=0, \lim_{x \to 1}t=1,  $$
and it turns out that when $x=1$, the integrand
$$ \frac{\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
   \sqrt{\frac{(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)}{2 u (x+1)}+1}}{u \sqrt{(1-u)
   (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}\bigg|_{x=1}=\frac{1}{\sqrt{2u(1-u)}}$$
</s>