Basic integration problem with log (1 Viewer)

Jeff_ · Nov 5, 2015

http://imgur.com/BT8Jqds

Anyone know why I get different answers when I do the u substitution (left) compared to just taking out the constant at the beginning? By the way the one on the right is correct...

This is a similar question to the one on https://www.youtube.com/watch?v=wt4Xvt6REFI and in the comments he explained that "ln(2x) + C = ln(2) + ln(x) + C = ln(x) + C' ", explaining that "the answer you get with the two techniques is only different by a constant" - I don't get what he means.

Thank you

Squar3root · Nov 5, 2015

both answers are correct

so the answer on the left you can go further to expand it out to be $\frac{1}{3}(ln(3) + ln(x)$ which is $\frac{1}{3}ln(3) + \frac{1}{3}ln(x)$

the $\frac{1}{3}ln(3)$ is just a constant and can be replaced with "+C"

InteGrand · Nov 5, 2015

Jeff_ said:
http://imgur.com/BT8Jqds

Anyone know why I get different answers when I do the u substitution (left) compared to just taking out the constant at the beginning? By the way the one on the right is correct...

This is a similar question to the one on https://www.youtube.com/watch?v=wt4Xvt6REFI and in the comments he explained that "ln(2x) + C = ln(2) + ln(x) + C = ln(x) + C' ", explaining that "the answer you get with the two techniques is only different by a constant" - I don't get what he means.

Thank you

$One of your answers is $\frac{1}{3}\ln \left| 3x\right|+C_1$. Another is $\frac{1}{3}\ln \left| x\right|+C_2$ ($C_1,C_2$ arbitrary constants). Note that the first answer an be simplified to $\frac{1}{3}\left(\ln \left|x\right| + \ln 3\right)+C_1 = \frac{1}{3} \ln |x| + \frac{1}{3}\ln 3 + C_1 = \frac{1}{3}\ln |x| + C$, where $C$ is just an arbitrary constant still. So your two answers just differ by a constant, so they are both antiderivatives of the function you integrated. It's like how $\sin x + 2 + c$ and $\sin x + c$ (which differ by a constant) are both answers to $\int \cos x \text{ d}x$.$

Jeff_ · Nov 5, 2015

Squar3root said:
both answers are correct

so the answer on the left you can go further to expand it out to be $\frac{1}{3}(ln(3) + ln(x)$ which is $\frac{1}{3}ln(3) + \frac{1}{3}ln(x)$

the $\frac{1}{3}ln(3)$ is just a constant and can be replaced with "+C"

InteGrand said:
$One of your answers is $\frac{1}{3}\ln \left| 3x\right|+C_1$. Another is $\frac{1}{3}\ln \left| x\right|+C_2$ ($C_1,C_2$ arbitrary constants). Note that the first answer an be simplified to $\frac{1}{3}\left(\ln \left|x\right| + \ln 3\right)+C_1 = \frac{1}{3} \ln |x| + \frac{1}{3}\ln 3 + C_1 = \frac{1}{3}\ln |x| + C$, where $C$ is just an arbitrary constant still. So your two answers just differ by a constant, so they are both antiderivatives of the function you integrated. It's like how $\sin x + 2 + c$ and $\sin x + c$ (which differ by a constant) are both answers to $\int \cos x \text{ d}x$.$

Thanks for your quick replies!

Forgot about the log rules (actually haven't really learned them yet as I joined my headstart tutoring class midway). Thanks

Basic integration problem with log (1 Viewer)

Jeff_

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Squar3root

realest nigga

InteGrand

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Jeff_

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