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Confused over basic integration - help !

Question:

(2x)1 dx\int (2x)^{-1}\ dx

Method 1

Take out the half and integrate normally, you will get:

0.5ln(x)0.5 ln (x)

Method 2

Sub in u = 2x and proceed, I get:

0.5ln(2x)0.5 ln (2x)


What ??
Reply 1
don't you have to make it 1/2x and then integrate normally? So, I think method 1's correct...
Reply 2
Original post by MV=PT
Question:

(2x)1 dx\int (2x)^{-1}\ dx

Method 1

Take out the half and integrate normally, you will get:

0.5ln(x)0.5 ln (x)

Method 2

Sub in u = 2x and proceed, I get:

0.5ln(2x)0.5 ln (2x)


What ??

Notice that 12ln2x=12ln2+12lnx\frac{1}{2}\ln 2x = \frac{1}{2}\ln 2 + \frac{1}{2}\ln x .

This is a just a constant (12ln2\frac{1}{2}\ln 2) different from your other answer so they both differentiate to give the original function.


The constant of integration can take any form. If you add 12lnc\frac{1}{2}\ln c to your first answer you get:

12lnx+12lnc=12lncx\frac{1}{2}\ln x + \frac{1}{2}\ln c = \frac{1}{2}\ln cx

So the second answer is just the case where c=2.
(edited 10 years ago)
Reply 3
Original post by notnek
Notice that 12ln2x=12ln2+12lnx\frac{1}{2}\ln 2x = \frac{1}{2}\ln 2 + \frac{1}{2}\ln x .

This is a just a constant (12ln2\frac{1}{2}\ln 2) different from your other answer so they both differentiate to give the original function.


The constant of integration can take any form. If you add 12lnc\frac{1}{2}\ln c to your first answer you get:

12lnx+12lnc=12lncx\frac{1}{2}\ln x + \frac{1}{2}\ln c = \frac{1}{2}\ln cx

So the second answer is just the case where c=2.


You have solved something very simple which has been troubling me fore about two years, but I never asked anyone :smile:

If I had to integrate (2x)-1 to use it as an integrating factor in a DE (so I will raise it to the integral power e) then the final answer I get as the solution to the equation will be different depending on whether I have ln 2x or ln x.
As eln 2x = 2x
but eln x = x
so I will be pretty much end up with different answers to my solution, but there can only be one correct general solution and both methods I have used to get there are correct ?
Reply 4
Original post by MV=PT
You have solved something very simple which has been troubling me fore about two years, but I never asked anyone :smile:

If I had to integrate (2x)-1 to use it as an integrating factor in a DE (so I will raise it to the integral power e) then the final answer I get as the solution to the equation will be different depending on whether I have ln 2x or ln x.
As eln 2x = 2x
but eln x = x
so I will be pretty much end up with different answers to my solution, but there can only be one correct general solution and both methods I have used to get there are correct ?


You multiply the whole equation by the integrating factor, so a constant multiplier won't make a difference - you can divide it back out again!

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