Anti derivatives: 1/(x^2)?

Whenever you're looking for the antiderivative of x^n, where n is ANY number whatsoever (be it a natural number, an integer, a fraction, or an irrational), there are only 2 cases you need to distinguish between.

If n=-1, you're integrating x^(-1) = 1/x, and its antiderivative is ln|x| + C.

If n is ANY number other than 1, then the antiderivative of x^n is (x^(n+1))/(n+1) + C.

So to integrate 1/x^2, you have 1/x^2 = x^(-2), so the antiderivative is (-1)*x^(-1) + C.

For it to be a natural log the function has to read either 1/x or (with u-substitution) 1/u, nothing else.

For your function, it's 1/x^2 so it's not a natural log.

You can take the x^2 to the top to read x^-2, then integrate.

You'd get -x^-1 + C, or -1/x +C

I recommend the following to avoid such confusion: 1/x can be rewritten as x^-1

Similarly, 1/(x^n) = x^-n

Anti-derivative for polynomial x^n = (x^n+1)/n+1

So using my recommendation, anti derivative of 1/(x^n) = anti derivative of x^-n = (x^-n+1)/-n+1

Similarly, anti derivative of 1/x = anti derivative of 1/(x^1) = anti derivative of x^-1 = x^(-1+1)/(-1+1) = x^0/0 = 1/0 = ERROR zero in the denominator. Hence, we know it's wrong.

But we can solve this using the definition of the derivative of the on function. And that's the definition you use to solve it.