BMAD 64 Report post 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 Share this post Link to post Share on other sites
0 rocdocmac 9 Report post 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 Share this post Link to post Share on other sites
1 rocdocmac 9 Report post Posted January 12, 2018 ∫1/(2x) = ½ln|x| + c. ln|2x| = ln|x| + ln(2), where ln(2) is a constant Quote Share this post Link to post Share on other sites
0 ThunderCloud 5 Report post Posted January 12, 2018 (edited) (removed... still thinking...) Edited January 12, 2018 by ThunderCloud Quote Share this post Link to post Share on other sites
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.
Share this post
Link to post
Share on other sites