Evaluating $\int_{-1}^2\frac{1}{x}dx$ I understand that when evaluating
$$ \int_{-1}^{2} \frac{1}{x} \mathrm dx = \ln 2$$
It's simple integration, I understand. I'm more focused on the theory behind if it even exists.  I had a question from Larson, Edward's Calculus 9th edition that was a true or false relating to this earlier today.
In one sense, I thought it had to be $\ln 2$. But, when looking at it again, technically the area from $-1$ to $0$, and $0$ to $1$ are negative infinity and infinity, respectively.  They should cancel out and our original integral from $-1$ to $2$ is the same as the integral from $1$ to $2$.  However, technically the integral does not converge from those two endpoints.  The areas from $-1$ to $0$ and $0$ to $1$ are only infinitesimally close as they both approach zero from one side.
I was looking for your guys' input here.  I was debating myself most of today over this seamlessly simple true/false question.
Note: this is a Calc 1 class, but I suppose I'm just getting ideas from Calc 2 onward (I've self-studied a bit) intertwined with our knowledge we've learned so far. 
 A: Just a small remark here, usually in introductory calculus classes, when we consider an integral of the form $$ \int_a^b f(x) dx$$ where $f(x)$ is discontinuous at the point $c$ but is otherwise continuous on the interval $[a,b]$, then we say that the integral exists if and only if both the improper integrals $\int_a^c f(x) dx$ and $\int_c^b f(x) dx$ exist and then we define
$$\int_a^b f(x) dx = \int_a^c f(x)dx + \int_c^bf(x)dx$$
As you have already observed, in your case with $f(x) = 1/x$, $a = -1$, $c = 0$, $b = 2$, neither of the improper integrals $\int_{-1}^0\frac{1}{x} dx$ or $\int_0^2 \frac{1}{x} dx$ are defined as their values would be "infinite". So by the usual Calculus definition, the integral you are considering doesn't exist!
I would suggest to you that this is a good way to think about this situation. In particular, the Fundamental Theorem of Calculus does not apply since $f(x)$ has a vertical asymptote (usually the FTC is only proved for either continuous functions or functions that have only a finite number of jump discontinuities). Therefore your initial calculation using FTC is incorrect.
A: What you are getting is known as the Cauchy Principal Value for this Improper Integral. It is actually
$$
\begin{align}
\lim_{\epsilon\to0^+}\left(\int_{-1}^{-\epsilon}\frac{1}{x}\mathrm{d}x+\int_{\epsilon}^{2}\frac{1}{x}\mathrm{d}x\right)
&=\lim_{\epsilon\to0^+}\left([\log(\epsilon)-\log(1)]+[\log(2)-\log(\epsilon)]\vphantom{\int}\right)\\
&=\lim_{\epsilon\to0^+}\left(\log(2)-\log(1)\vphantom{\int}\right)\\
&=\log(2)
\end{align}
$$
