Question

What is the antiderivative of `(1/(x^2))`?

Answer 1

`-1/x+C`

Explanation:

`d/dx(x^r+C) = rx^(r-1)` (for constant `C`)

So, for all `s != 0`, the antiderivative of `x^s` is `x^(s+1)/(s+1)+C`.

`1/x^2 = x^-2` so its antiderivative is `x^(-2+1)/(-2+1) = x^-1/-1 = -1/x` don't forget the `+C`.