You can understand it in terms of infinite series (addition rather than integration) if you are more familiar with those ideas. In particular, you can estimate the integral of 1/x² in terms of

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ... 

(view this as a Riemann sum, for Δx = 1, draw the picture, you'll see that this is an underestimate for the integral. Similarly,

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...

is an overestimate, so the value of the integral is between the two).

This infinite series converges (the sum has a finite value). You can informally explain this by considering a piece of paper; cut it in half, save one half, cut the other half in half again. Save one of these quarters, cut the other quarter in half again. Save one of the resulting eighths, cut the other eighth in half, etc. Each new cut corresponds to another term of the series. It is easy to reason in this case, that the sum of all the pieces is 1 piece of paper.

In the case of 1/x, we look at the series

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...

(this will for sure be an underestimate for the integral, so if the series is divergent, so is the integral).

You can reason in the following way to show this diverges: you know 1/3 > 1/4, you know, 1/5, 1/6, 1/7 all > 1/8, you know 1/9, 1/10, 1/11, 1/12, 1/13, 1/14, 1/15 all > 1/16, etc. So, you can write:

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... > 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16 + 1/16 + ....) 
= 1/2 + 1/2 + 1/2 + 1/2 + ...

which clearly diverges.

Basically, while the terms in both series (and corresponding integrals) get smaller and smaller, 1/x² gets smaller much faster than 1/x, and this makes a huge difference if you try and add all those up. 1/x is in fact the line of demarcation for convergence like this: if you have an exponent in the denominator of 1 or less, you're not going to converge, if it is any higher than 1, it will. (eg 1/x^0.9, 1/x do not converge, 1/x^1.1 and 1/x²⁵ do converge)