There IS an antiderivative (integral) of 1/x. It is the natural log-base-e (ln) function.
In your case
f'(x) = -1/x + C1
f"(x) = - ln x + C1 x + C2
where C1 and C2 are arbitrary constants
Can someone help me with this problem? Find f if f''(x) = x^-2, x>0, f(1)=0, and f(2)=0.
I tried using antiderivatives but I realized I couldn't take the antiderivative of x^-1.
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1 answer