2007-07-02 05:52:00 · 7 answers · asked by jennifer 1 in Science & Mathematics ➔ Mathematics
I'ts -ln|16-x| because of the negative coefficient of x.
This can also be rewritten using log laws as
+ln|1/(16-x)|
2007-07-02 06:03:33 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋
It's -ln(16-x).
Try and differentiate the result. You need to use the chain rule which produces a -1 from the -x within the equation.
2007-07-02 13:01:10 · answer #2 · answered by Richie 2 · 0⤊ 0⤋
Lets substitute 16-x = t. Therefore differentiating both sides of 16-x = t we get => -dx = dt or dx = -dt.
Substituting everything in to the integral we get,
Integral(dx/(16-x)) = Integral(-dt/t) = - Integral(dt/t) = -ln(t).
substituting back 16-x = t, we get the answer to be -ln(16-x).
2007-07-02 12:59:54 · answer #3 · answered by ping_anand 3 · 0⤊ 0⤋
-ln(16-x) because the derivative of a ln is the derivative of the function (16-x) over the function.
2007-07-02 13:00:54 · answer #4 · answered by therealchuckbales 5 · 0⤊ 0⤋
1/(16-x) dx {1/(1-x/b) dx = bln(1-x/b)}
1/16*{1/(1-x/16)}dx
1/16*16*ln(1-x/16)
ln(1-x/16)
2007-07-02 13:04:18 · answer #5 · answered by Anonymous · 0⤊ 0⤋
I = â« 1 / (16 - x) dx
Let u = 16 - x
du / dx = - 1
dx = - du
I = - â« (1 / u) du
I = - log u + C
I = - log (16 - x) + C
2007-07-06 04:10:04 · answer #6 · answered by Como 7 · 0⤊ 0⤋
Math is hard
2007-07-02 12:58:48 · answer #7 · answered by Anonymous · 0⤊ 0⤋