Question

f'(x) = (x^2-1)/x, f(1) = .5 and f(-1) = 0
Find f(x)

Answer 1

\[f'(x)=\frac{x^2-1}{x}=x-\frac{1}{x}\] so
\[f(x)=\frac{x^2}{2}-\ln(x)+C\] and there is no way that i can see that
\[f(-1)\] exists. what am i missing?

Answer 2

I thought the same thing but the textbook gives both of those values.

Answer 3

\[f(x)=x ^{3}/3+lnx+c\]

Answer 4

Actually, isn't the integral of 1/x ln|x|?