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Integrating 1/(10-8x)dx

t =∫110−8xdx=−18ln(10−8x)+c \int \frac{1}{10-8x}dx = -\frac{1}{8}ln(10-8x) + c . However book says it equals ∫−18ln(5−4x)+c \int -\frac{1}{8} ln(5-4x) + c . I can see they've taken a common factor of 2 out but I don't understand where its gone. And also when I evaluate it with t=0, x=0 I would get a different c value.
Reply 1
Avatar for Spungo
Spungo
10
What would (-1/8) ln(5 - 4x) + (-1/8) ln(2) be? What is the difference in your value of C with theirs?
(edited 12 years ago)
Reply 2
Avatar for Freerider101
Freerider101
OP
12
Original post by Spungo
What would (-1/8) ln(5 - 4x) + (-1/8) ln(2) be? What is the difference in your value of C with theirs?


I've just realised I should of taken a half out as a common factor before I integrated. I'm still confused as to why I get a different answer though.
Reply 3
Avatar for Freerider101
Freerider101
OP
12
Also ∫12xdx=12∫1xdx=12lnx+c \int \frac{1}{2x} dx = \frac{1}{2} \int \frac{1}{x} dx = \frac{1}{2}lnx + c . And again if I kept the 2 in there I would of got 12ln2x+c \frac{1}{2}ln2x + c which is wrong. Why is this?
Original post by Freerider101
t =∫110−8xdx=−18ln(10−8x)+c \int \frac{1}{10-8x}dx = -\frac{1}{8}ln(10-8x) + c . However book says it equals ∫−18ln(5−4x)+c \int -\frac{1}{8} ln(5-4x) + c . I can see they've taken a common factor of 2 out but I don't understand where its gone. And also when I evaluate it with t=0, x=0 I would get a different c value.

As Spungo said, the only difference is in the arbitrary constant of integration. Note that:
−18ln⁡(10−8x)+c=−18ln⁡(2(5−4x))+c=−18ln⁡(5−4x)+(−18ln⁡2+c)=−18ln⁡(5−4x)+K-\dfrac{1}{8}\ln (10-8x) + c = -\dfrac{1}{8} \ln (2(5-4x)) + c = -\dfrac{1}{8}\ln (5-4x) +(-\dfrac{1}{8}\ln 2 + c) = -\dfrac{1}{8}\ln (5-4x)+K
Reply 5
Avatar for steve2005
steve2005
17
Original post by Freerider101
t =∫110−8xdx=−18ln(10−8x)+c \int \frac{1}{10-8x}dx = -\frac{1}{8}ln(10-8x) + c . However book says it equals ∫−18ln(5−4x)+c \int -\frac{1}{8} ln(5-4x) + c . I can see they've taken a common factor of 2 out but I don't understand where its gone. And also when I evaluate it with t=0, x=0 I would get a different c value.


Can you give the original question? If it's from a past paper can you tell us which one?
Reply 6
Avatar for Spungo
Spungo
10
Original post by Freerider101
I've just realised I should of taken a half out as a common factor before I integrated. I'm still confused as to why I get a different answer though.


The "c" is an arbitrary constant.
The difference comes because of the slightly different way of writing the function, but they are both the same thing really.

If A is a constant, then A + ln(2) is still a constant and could just be written as B.
Reply 7
Avatar for rub em out
rub em out
7
Original post by Freerider101
t =∫110−8xdx=−18ln(10−8x)+c \int \frac{1}{10-8x}dx = -\frac{1}{8}ln(10-8x) + c . However book says it equals ∫−18ln(5−4x)+c \int -\frac{1}{8} ln(5-4x) + c . I can see they've taken a common factor of 2 out but I don't understand where its gone. And also when I evaluate it with t=0, x=0 I would get a different c value.


Pretty sure your book is wrong.. :/ as far as I can see you are correct

edit: actually they are both the same thing, 1/2(5-4x) = 1/4 * 2ln(5-4x) = 1/8ln(5-4x)
(edited 12 years ago)
Original post by rub em out
Pretty sure your book is wrong.. :/ as far as I can see you are correct

Neither the book, nor he, are wrong. They're just two different ways of expressing the same result.
Reply 9
Avatar for Freerider101
Freerider101
OP
12
Original post by Spungo
What would (-1/8) ln(5 - 4x) + (-1/8) ln(2) be? What is the difference in your value of C with theirs?


Yep I get 18ln5 {1}{8}ln5 as the time using both methods.
Original post by Freerider101
Also ∫12xdx=12∫1xdx=12lnx+c \int \frac{1}{2x} dx = \frac{1}{2} \int \frac{1}{x} dx = \frac{1}{2}lnx + c . And again if I kept the 2 in there I would of got 12ln2x+c \frac{1}{2}ln2x + c which is wrong. Why is this?


Original post by Farhan.Hanif93
As Spungo said, the only difference is in the arbitrary constant of integration. Note that:
−18ln⁡(10−8x)+c=−18ln⁡(2(5−4x))+c=−18ln⁡(5−4x)+(−18ln⁡2+c)=−18ln⁡(5−4x)+K-\dfrac{1}{8}\ln (10-8x) + c = -\dfrac{1}{8} \ln (2(5-4x)) + c = -\dfrac{1}{8}\ln (5-4x) +(-\dfrac{1}{8}\ln 2 + c) = -\dfrac{1}{8}\ln (5-4x)+K


Thanks a lot for that. So in my above quote, are both answers acceptable?

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