## Mertsj 3 years ago f'(x) = (x^2-1)/x, f(1) = .5 and f(-1) = 0 Find f(x)

1. satellite73

$f'(x)=\frac{x^2-1}{x}=x-\frac{1}{x}$ 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?

2. Mertsj

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

3. jerwyn_gayo

$f(x)=x ^{3}/3+lnx+c$

4. Mertsj

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

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