BMAD 63 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. Share this post Link to post Share on other sites

0 rocdocmac 8 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 Share this post Link to post Share on other sites

1 rocdocmac 8 Report post Posted January 12, 2018 ∫1/(2x) = ½ln|x| + c. ln|2x| = ln|x| + ln(2), where ln(2) is a constant 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 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.

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