is the integral of 1/x^2, 2x ln of the absolute value of x^2?
no, the integral of dx/x^2 = -1/x + constant
You only have to do the ln thing if it is the integral of 1/x
otherwise
integral x^n dx = (1/n+1) x^(n+1)
that works fine for n = -2
integral x^-2 dx = (1/-1) x^-1 as I said
HOWEVER in the nasty case of
integral x^-1 dx = (1/0) x^0
BUT
1/0 is UNDEFINED
thnx
To find the integral of 1/x^2, you can use the power rule of integration.
The power rule states that the integral of x^n, where n is any real number except -1, is equal to (x^(n+1))/(n+1), plus a constant of integration.
In this case, the integral of 1/x^2 can be found by applying the power rule with n=-2:
∫(1/x^2) dx = (x^(-2+1))/(-2+1) + C
= x^(-1)/(-1) + C
= -1/x + C
Therefore, the integral of 1/x^2 is -1/x + C, where C is the constant of integration.
The expression you mentioned, 2x ln| x^2|, is not the correct antiderivative of 1/x^2.