2006-12-21 13:40:34 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
ln|x| + C
Because you can't have gotten 1^-x from x^0 because that would have been one. This is one of those Derivitive/Integral combos you have to memorize, like sinx, cosx, etc.
2006-12-21 13:44:28 · answer #1 · answered by toothpickgurl 3 · 0⤊ 0⤋
One thing you need to note when taking the antiderivative of x^(-1) is that it doesn't follow the typical reverse power rule. For any other power, if f(x) = x^n, then the antiderivative, F(x) is equal to
F(x) = [x^(n+1)]/[n+1] + C
But if n = -1, then you have to remember that the antiderivative is the natural log, or ln, of the ABSOLUTE value of x.
f(x) = 1/x
F(x) = ln|x| + C
2006-12-21 22:39:03 · answer #2 · answered by Puggy 7 · 0⤊ 0⤋
antiderivative of x^-1 = log x + c
2006-12-21 21:46:55 · answer #3 · answered by Kinu Sharma 2 · 0⤊ 0⤋
ln|x| + c
Don't forget the absolute value. ln x is undefined when x is negative.
2006-12-21 22:21:43 · answer #4 · answered by sahsjing 7 · 2⤊ 0⤋
ln x
2006-12-21 21:41:29 · answer #5 · answered by Modus Operandi 6 · 0⤊ 0⤋
y = lnx --->
x = e^y
implicit differentiation:
1 = e^y * dy/dx
1 = x * dy/dx
1/x = dy/dx.
2006-12-21 21:50:14 · answer #6 · answered by Anonymous · 0⤊ 0⤋
ln(x) + C, where C is a constant.
2006-12-21 21:43:34 · answer #7 · answered by Northstar 7 · 0⤊ 0⤋
lnx+C
2006-12-21 22:04:28 · answer #8 · answered by raj 7 · 0⤊ 0⤋