What is the antiderivative of f'(x) = (1) / (x+2)?
If you must know my *** work... (in an attempt to use the formula that I got from a Youtube video x^n dx = [x^(n+1) / (n+1)] + C)
f'(x) = (1) / (x+2)
f'(x) = (x+2)^(-1)
f(x) = (x+2)^(-1+1) / (-1+1)
f(x) = ...undefined
And the answer is f(x) = ln(x+2) + C
Can someone please explain to the idiot I am?
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Verified answer
The answer is ln|x+2|+C. You probably just forgot that the derivative of the natural log is 1/x. Remember that the integral of 1 divided by anything is ln(that); plus chain rule if it applies.
f'=1/(x+2)
f(x) = ln (x+2) +c
ln(x+2) + C
f (x) = log [ x + 2 ] + C