An explanation of the integration

Question

So, the integral is: $$\int_1^2\frac{x-2}{\sqrt{x-1}}dx$$ If I copied correctly from the board, the teacher said if x approaches 1+, the function approaches
+$\infty$. What is the difference between 1- and 1+? If x approaches 1+, isn't the function approaching -$\infty$, not +$\infty$.
One more question: When solving that integral, the teacher wrote: $$\int_1^2\frac{x-1-1}{\sqrt{x-1}}dx=\int_1^2\frac{x-1}{\sqrt{x-1}}dx-\int_1^2\frac{1}{\sqrt{x-1}}dx=$$$$\int_1^2\sqrt{x-1}dx-\int_1^2\frac{1}{\sqrt{x-1}}dx=I1+I2$$ For $I1$, we substitute $x-1$ with $z$ and when we solve, we get $I1=\frac{2}{3}$
For I2, the teacher wrote: $$I2 = \lim_{ɛ->0}\int_{1+ɛ}^2\frac{1}{\sqrt{x-1}}dx$$ And when we substitute $x-1$ with $z$ again, we solve $I2$ and get $I2=1$. My question is, why did the teacher change the integration from $1$ to $2$, to $1+ɛ$ to $2$, what does that $ɛ$ represent?

Answer 1

The function $ y=\dfrac{1}{\sqrt{x-1}}$ is not defined for $x=1$, so your second integral is an improper integral that can be evaluated as a limit $$ \lim_{t\rightarrow 1^+}\int_{t}^2\frac{1}{\sqrt{x-1}}dx $$ and the integral exists if this limit exists and is finite.

Writing $1+\epsilon $ for $\epsilon \rightarrow 0$ is the same as $t=1+\epsilon \rightarrow 1$.