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C4 Integration - Logarithms Question

It's given that ∫(1/x) is Ln(|x|). For example, ∫(1/(x+1) is Ln(x+1).

But given this, how is ∫(1/3x+2) NOT Ln(3x+2), but instead 1/3(Ln(3x+2))? My book just goes on to say that this is due to the chain rule but does not specify the reasoning at all. Thanks.
(edited 5 years ago)
Let u=3x+2, bring the constant out the front.
1/3*(integral of 1/u)
=1/3*ln(u)
=1/3*ln(3x+2)

EDIT: https://i.imgur.com/sq6ekGW.jpg
(edited 5 years ago)
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Dggj_19
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Original post by sdr981
Let u=3x+2, bring the constant out the front.
1/3*(integral of 1/u)
=1/3*ln(u)
=1/3*ln(3x+2)

If you're taking that constant out of the integral, that implies that before you do it's ∫(1/3u), right? The only way I can see you being able to bring the coefficient out is if u is (x+2/3), so 3u is 3x+2, which then lets you take the 1/3 out of the integral.
Original post by Dggj_19
It's given that ∫(1/x) is Ln(|x|). For example, ∫(1/(x+1) is Ln(x+1).

But given this, how is ∫(1/3x+2) NOT Ln(3x+2), but instead 1/3(Ln(3x+2))? My book just goes on to say that this is due to the chain rule but does not specify the reasoning at all. Thanks.


If you've learnt integration by substitution, then just sub in u=3x+2u = 3x+2 and see for yourself.
(edited 5 years ago)
Original post by Dggj_19
If you're taking that constant out of the integral, that implies that before you do it's ∫(1/3u), right? The only way I can see you being able to bring the coefficient out is if u is (x+2/3), so 3u is 3x+2, which then lets you take the 1/3 out of the integral.

Sorry for my lack of clarity. I am attaching my workings (hopefully you can read them :smile:):
https://i.imgur.com/sq6ekGW.jpg

Essentially, the integral is still with respect to dx after the u-sub, so we must find du/dx and then replace dx with some function of du.
(edited 5 years ago)

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