$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$ 
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 ?
 A: I found a possible method here which is straightforward.  I will just carry it here.
\begin{aligned}
\lim _{x \rightarrow 1} u(x) &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t} \cdot \frac{1+t}{(t-1)^{2}}}{\sqrt{\frac{1+x}{(x-1)^{2}}-\frac{1+t}{(t-1)^{2}}}} d t \\
&=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t}(1+t)(1-x)}{(1-t) \sqrt{(x-t)(3-t-x-x t)}} d t \\
&=\lim _{y \rightarrow 0} \int_{y}^{1} \frac{(2-s) \sqrt{1-s} y}{s \sqrt{(s-y)(2 s+2 y-s y)}} d s \quad(\text { Let } s=1-t, y=1-x)
\\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y} \cdot y \cdot y}{u y \sqrt{(u y-y)\left(2 u y+2 y-2 u y^{2}\right)}} d u \quad(\text { Let } s=u y) \\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y}}{u \sqrt{(u-1)(2 u+2-u y)}} d u\\
&=\sqrt{2} \int_{1}^{+\infty} \frac{1}{u \sqrt{u^{2}-1}} d u \\
&=\sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-m^{2}}} d m \quad\left(\text { Let } m=\frac{1}{u}\right) \\
&=\frac{\sqrt{2}}{2} \pi
\end{aligned}
A: 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
$$ \frac{\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
   \sqrt{\frac{(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)}{2 u (x+1)}+1}}{u \sqrt{(1-u)
   (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}\bigg|_{x=1}=\frac{1}{\sqrt{2u(1-u)}}$$
A: I'm somehow not really satisfied with all those great solutions above. To me, they didn't show quite well the underlying reason that makes that convergence occur. So I'll give another solution by trying to generalize the very first problem.
I'll start by stating something trivial.
Lemma 1: (Convergence of measures)
Let $\mu_1, \mu_2,...$ be a sequence of measures on $[0,1]$ such that $$ \lim_{ n \rightarrow \infty} \mu_n[0,1-\epsilon) = 0 $$
for all $\epsilon>0$, and $$ \lim_n \mu_n [0,1]=C$$
for some $C>0$, then for all continuous function $g$ on $[0,1]$ , we have:
$$ \int g d\mu_n \longrightarrow Cg(1)$$
$\square$
Now, let's go back to the initial question.
Main solution
I'll define:

*

*$g(t):= \frac{ \sqrt{2t} f(t)^{3/2}}{f'(t)}$

*$d\mu_x(t)= \mathbb{1}_{[0,x]}(t)\frac{1}{ \sqrt{\left(1- h_x(t)\right)h_x(t)}} h_x'(t)dt $, where $h_x(t)= \frac{f(t)}{f(x)}$
So we see that:

*

*The integral whose limit we want to calculate is equal to $ \int g d\mu_x$

*$g$ is continuous on $[0,1]$ and  $g(1)=\frac{\sqrt{2}}{2}$

*Note that $h_x$ is increasing in $t$, thus by changing variables, we can show that $$\mu_x[0,1] = \int_{[0,1]} \frac{1}{\sqrt{u(1-u)}} du= \pi$$

*Also by changing variables and noting that $\lim_{x \rightarrow 1^-} f(x)=+\infty$, we can show that:
$$\lim_{x \rightarrow 1^{-}}  \mu_x[0,\alpha] =0$$ for all $\alpha \in (0,1)$
Hence by applying the very first lemma, we have the final result
Comment:
TL;DR: the main idea is just to find the right $g$ and $\mu_x$ to compactify our integral.
