Hello, I was doing the antiderivative of: f'(x) = (x^2-1)/x Am I allowed to separate this into x^2/x - 1/x? Or do the brackets prevent me from doing so? If not, how do I approach this antiderivative? Thanks for anyone's help!
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Originally Posted by ty2391 Hello, I was doing the antiderivative of: f'(x) = (x^2-1)/x Am I allowed to separate this into x^2/x - 1/x? Or do the brackets prevent me from doing so? If not, how do I approach this antiderivative? Thanks for anyone's help! you have the correct approach ... $\displaystyle \frac{x^2-1}{x} = x - \frac{1}{x}$ antiderivative is $\displaystyle \frac{x^2}{2} - \ln|x| + C$
Originally Posted by skeeter you have the correct approach ... $\displaystyle \frac{x^2-1}{x} = x - \frac{1}{x}$ antiderivative is $\displaystyle \frac{x^2}{2} - \ln|x| + C$ Thanks a lot, I forgot to mention that f(1) = 1/2, and f(-1) = 0 When I solve for c, I get two different values when x=1 and x=-1. (C=0 when x=1, C=1/2 when x=-1) Does this mean C=0 when x>0 and C=1/2 when x<0?
Originally Posted by ty2391 Thanks a lot, I forgot to mention that f(1) = 1/2, and f(-1) = 0 When I solve for c, I get two different values when x=1 and x=-1. (C=0 when x=1, C=1/2 when x=-1) Does this mean C=0 when x>0 and C=1/2 when x<0? looks that way, doesn't it?
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