Question on improper integrals $$\int_0^2 \dfrac{\mathrm dx}{\sqrt{x}(x-1)}$$
I want to determine whether this integral converges or diverges. Now usually problems like these are easy, but this one is kind of tricky since it is discontinuous at both 0 and 1. Whereas with one situations where the integral is only discontinuous at 1 value, I could just set up 2 integrals and use the "lim of t" method, but here I cant set up two integrals. The only way that two integrals could be set up is if the first one is from 0 to 1 and the second one is from 1 to 2, but the first one is discontinuous at both values, so not sure what the work around is.  
 A: You can break up the integral at any point or points you like. In this case, you could break it up into three integrals: pick a point $c$ strictly between $0$ and $1$, and consider:
\begin{align*}
\int_0^2 \frac{dx}{\sqrt{x}(x-1)} &= \int_0^c\frac{dx}{\sqrt{x}(x-1)}+\int_c^1\frac{dx}{\sqrt{x}(x-1)} + \int_1^2\frac{dx}{\sqrt{x}(x-1)}\\
&= \lim_{a\to 0^+}\int_a^c \frac{dx}{\sqrt{x}(x-1)} + \lim_{b\to 1^-}\int_c^b\frac{dx}{\sqrt{x}(x-1)} + \lim_{d\to 1^+}\int_d^2\frac{dx}{\sqrt{x}(x-1)}.
\end{align*}
The original improper integral exists if and only if each of the three improper integrals exist.
A: Consider the change of variables $t=\sqrt x$. It transforms the interval $[0,2]$ into $[0,\sqrt 2]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{{\rm e}^{#1}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\sgn}{{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}$Try to solve the integral. You'll find whether it converges. For example,
$\ds{\int_{\epsilon}{\dd x \over \sqrt{x\,}\,\pars{1 - x}}}$ with $\epsilon > 0$.
Later on, you can study the limit $\ds{\epsilon \to 0^{+}}$. In this way, you "kill two birds with a one shot".
