Convergence of $\int_{1^+}^{3^-}1/\sqrt{(3-x)(x-1)}dx$ I'm trying to study the convergence of 
$$\int_{1^+}^{3^-}\frac{1}{\sqrt{(3-x)(x-1)}}dx$$
But I'm not sure how to get started. Any suggestion or tip, for this kind of convergence problem?
The indefinite integral seems a bit a of a mess, involving the hyperbolic sine.
Thanks in advance!
 A: The integral is convergent since $x=1$ and $x=3$ are singularities of the integrand function, but integrable singularities:
$$ \int_{0}^{1}\frac{dx}{\sqrt{x}} \stackrel{x=t^2}{=} \int_{0}^{1}\frac{2t\,dt}{t} = 2.\tag{1}$$
In our case:
$$ \int_{1}^{3}\frac{dx}{\sqrt{(x-1)(3-x)}}\stackrel{x=t+2}{=}\int_{-1}^{1}\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)|_{-1}^{1}=\color{red}{\pi}.\tag{2} $$
A: $$I=\int_{1^+}^{3^-}\frac{1}{\sqrt{(3-x)(x-1)}}dx=\int_{1^+}^{3^-}\frac{1}{\sqrt{1-(x-2)^2}}dx=\int_{-1^+}^{1^-}\frac{1}{\sqrt{1-x^2}}dx$$
we have
$$I=\sin^{-1}(x)\Big{|}_{-1^+}^{1^-}=\pi$$
A: If you just want convergence, then compare the integral with appropriate $p$-integrals.  Near $x=1$, $1/\sqrt{3-x}$ is less than $1$, so
$$\int_{1+}^{2} \frac{1}{\sqrt{3-x}\sqrt{x-1}} \; dx <\int_{1+}^{2} \frac{1}{\sqrt{x-1}} \; dx$$ which converges.
Likewise, near $x=3$, $1/\sqrt{x-1}$ is less than one.  So $$\int_{2}^{3-} \frac{1}{\sqrt{3-x}\sqrt{x-1}} \; dx <\int_{2}^{3-} \frac{1}{\sqrt{3-x}} \; dx$$ which converges.  Then add the pieces together.
A: Let $y=\frac{x-1}{3-1}$, then
\begin{align}
\int\limits_{1}^{3} \frac{1}{\sqrt{(3-x)(x-1)}} \mathrm{d} x & = \int\limits_{0}^{1} (1-y)^{-1/2} y ^{-1/2} \mathrm{d} y \\
& = B\left(\frac{1}{2},\frac{1}{2}\right) \\
& = \pi
\end{align}
A: $$\text{I}=\int_{1^+}^{3^-}\frac{1}{\sqrt{(3-x)(x-1)}}\space\text{d}x=\int_{1^+}^{3^-}\frac{1}{\sqrt{1-(x-2)^2}}\space\text{d}x$$
Now, subsitute $u=x-2$ and $\text{d}u=\text{d}x$:
$$\text{I}=\int_{1^+-2}^{3^--2}\frac{1}{\sqrt{1-u^2}}\space\text{d}u$$
Now, the integral of $\frac{1}{\sqrt{1-u^2}}$ is the inverse sin:
$$\text{I}=\left[\arcsin(u)\right]_{1^+-2}^{3^--2}=\arcsin(3^--2)-\arcsin(1^+-2)=\arcsin(1)-\arcsin(-1)=\pi$$
Evaluate:


*

*$$\lim_{x\to3^-}\int\frac{1}{\sqrt{(3-x)(x-1)}}\space\text{d}x=0$$

*$$\lim_{x\to1^+}\int\frac{1}{\sqrt{(3-x)(x-1)}}\space\text{d}x=-\pi$$

