# Question on inappropriate integrals

$$\int_0^2 \dfrac{\mathrm dx}{\sqrt{x}(x-1)}$$

I intend to establish whether this indispensable converges or deviates. Currently generally troubles like these are very easy, yet this set is sort of complicated given that it is alternate at both 0 and also 1. Whereas with one scenarios where the indispensable is just alternate at 1 value, I can simply set up 2 integrals and also make use of the "lim of t" method, yet below I angle set up 2 integrals. The only manner in which 2 integrals can be set up is if the first one is from 0 to 1 and also the 2nd one is from 1 to 2, yet the first one is alternate at both values, so not exactly sure what the job about is.

Consider the adjustment of variables $t=\sqrt x$. It changes the interval $[0,2]$ right into $[0,\sqrt 2]$ and also gets rid of among the trouble absolutely nos. It is very easy to see from the resulting expression that the indispensable diverges.

You can separate the indispensable at *any kind of * factor or factors you such as. In this instance, you can damage it up right into *3 * integrals : select a factor $c$ purely in between $0$ and also $1$, and also take into consideration :
\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 initial inappropriate indispensable exists if and also just if each of the 3 inappropriate integrals exist.

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