is the integral of 1/x^2, 2x ln of the absolute value of x^2?
3 answers
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
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