Integral of odd function doesn't converge? When I look up $$\int_{-1}^1 \dfrac{1}{x} dx$$ 

on Wolfram Alpha, it says it doesn't converge. While this is a sum of two diverging integrals, the two areas are clearly symmetric, and I'd assume the answer would be zero. How does one usually treat integrals like this?
It's undefined quite simply because you can force it to have another value. First of all, let's say by $\int_{-1}^1\frac1xdx$ you mean
$$
\lim_{s \to 0^-\,\, t \to 0^+}\left(\int_{-1}^s\frac1x dx + \int_t^1 \frac1xdx\right)
$$
Now, if we force $s = -t$, then as you say, this evaluates to $0$. However, if we say $s = -at$ for some $a &gt; 0$, then we get something completely different (we get $\ln a$). It's this "completely different" that in the end makes the value indeterminable.



Side note: It's possible to go into subdividing the interval $[-1, 1]$ unevenly and use the definition of (Riemann) integral to see that we can get the sum to be whatever we want, and once again, this means undefined.
The integral is undefined. However consider the following formula:

$$lim_{\epsilon \to 0} \frac {x}{x^2+ \epsilon^2} = \frac {1}{x}$$

Let us try this representation of $1/x$ in your integral. Now perform the integration before taking the limit of $\epsilon$ to $0$. [This change-of-order is presumably not strictly allowed, but obviously it makes a lot of sense to do so anyway.] Note that the integrand is now divergence-free, so the integration is completely straightforward. The anti-derivative is found to be: 

$$(1/2) * log (x^2 + \epsilon^2)$$

This function is symmetric, hence the cancellation that we seek indeed takes place and the integral from $-1$ to $+1$ equals zero. Of course it depends on the context of the mathematical problem at hand whether a method like this is justified.