why does the integral of 1/x² dx from -1 to 1 equal -1?

Let me first answer your second question.

You integrate between your limits and discontinuities and

sum up those integrals.

For example your function has discontinuity at x = 0.

Hence you will have two integrals one from -1 to 0

and second from 0 to 1, Therefore your first question is:

limit(when e -> 0)

integral(from -1 to -e)[dx/x^2] + integral(from e to 1)[dx/x^2] =

(from -1 to -e)[-1/x] + (from +e to 1)[-1/x] = 1/e + 1 + (-1 + 1/e) = 2/e -> inf.

Therefore the -1 result is not correct.

It doesn't.

An integral can only be applied over a domain if the function is continuous along that domain.

The best way to integrate 1/x^2 is to first notice that 1/x^2 is an even function (f(-x) = f(x))

So if you're integrating from -a to a, it'd be just as easy to integrate from 0 to a and then double that

int(dx / x^2 , x = 0 , x = 1) =>

(-1/x) {0 , 1} =>

-1/1 - (-1/0) =>

1/0 - 1

Now, as x approaches 0 from the right, 1/x goes to infinity

inf - 1

inf

Multiply it by 2

infinity