BMAD Posted January 11, 2018 Report Share Posted January 11, 2018 If the antiderivative of u^-1 = ln |u| + c then why does this not follow: integrate 1/(2x) dx set u =2x, then du = 2 dx, then dx = (1/2)du Then we could integrate (1/2)(1/u)du By the definition above we get 1/2 ln|u| + c which means that the integration of 1/(2x) = 1/2 ln|2x|| + c However, this is a false statement. Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted January 13, 2018 Report Share Posted January 13, 2018 On 1/12/2018 at 11:53 AM, rocdocmac said: ∫1/(2x) = ½ln|x| + c Spoiler ln|2x| = ln|x| + ln(2), where ln(2) is a constant Thus ... "½ln|2x|" + c = ½(ln|x| + ln(2)) + c1 = ½ln|x| + c2 +c1 = ½ln|x| + c Quote Link to comment Share on other sites More sharing options...

1 rocdocmac Posted January 12, 2018 Report Share Posted January 12, 2018 ∫1/(2x) = ½ln|x| + c. ln|2x| = ln|x| + ln(2), where ln(2) is a constant Quote Link to comment Share on other sites More sharing options...

0 ThunderCloud Posted January 12, 2018 Report Share Posted January 12, 2018 (edited) (removed... still thinking...) Edited January 12, 2018 by ThunderCloud Quote Link to comment Share on other sites More sharing options...

## Question

## BMAD

If the antiderivative of u^-1 = ln |u| + c then why does this not follow:

integrate 1/(2x) dx

set u =2x, then du = 2 dx, then dx = (1/2)du

Then we could integrate

(1/2)(1/u)du

By the definition above we get

1/2 ln|u| + c

which means that the integration of 1/(2x) = 1/2 ln|2x|| + c

However, this is a false statement.

## Link to comment

## Share on other sites

## 3 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.