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Integrating partial fractions - Please Help

Hi all,

Sorry for posting this but I'm really stuck on a question. I know the method to do it but every time I try it, I cant get the answer in the form the question is asking!

The question is

Show that:
∫(Between 0 and -1) [5-8x] / [2+x][1-3x] dx = p ln2 , where p is rational

--> Splitting into partial fractions gives (3 / 2+x) + (1 / 1-3x)

I keep getting ln2 + ln(1/4), but that doesn't allow me to give my answer in the format they want.

Any help for this question would be MUCH appreciated :smile:
Thanks!
(edited 9 years ago)
Original post by MatthewP01

I keep getting ln2 + ln(1/4), but that doesn't allow me to give my answer in the format they want.

Any help for this question would be MUCH appreciated :smile:
Thanks!


You need to do something with that ln(1/4). Try expressing 1/4 as a power of 2, and hopefully you'll see what to do - remember your laws of logs.

Edit: Though I am puzzled as to how you end up with ln(1/4)
(edited 9 years ago)
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MatthewP01
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Original post by ghostwalker
You need to do something with that ln(1/4). Try expressing 1/4 as a power of 2, and hopefully you'll see what to do - remember your laws of logs.

Edit: Though I am puzzled as to how you end up with ln(1/4)


When you integrate the partial fraction 1 / 1-3x and add in the limits do you not get (ln1 - ln4). Therefore by rule of logs this equals ln(1/4)?

I've tried it again and got 3(ln2) + (ln1/4) and so I'm still kind of stuck :frown:
Original post by MatthewP01
When you integrate the partial fraction 1 / 1-3x and add in the limits do you not get (ln1 - ln4). Therefore by rule of logs this equals ln(1/4)?

I've tried it again and got 3(ln2) + (ln1/4) and so I'm still kind of stuck :frown:


What did you get when you integrated - prior to putting the limits in
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MatthewP01
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Original post by TenOfThem
What did you get when you integrated - prior to putting the limits in


Hi :smile:

I got 3[ln |2+x|](limits) + [ln |1-3x|](limits)

I have a feeling I'm going about this all wrong :frown:
Original post by MatthewP01
Hi :smile:

I got 3[ln |2+x|](limits) + [ln |1-3x|](limits)

I have a feeling I'm going about this all wrong :frown:


If you differentiated ln|1-3x| you would get −31−3x\dfrac{-3}{1-3x}


So you need to consider the inverse chain rule there
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MatthewP01
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Original post by TenOfThem
If you differentiated ln|1-3x| you would get −31−3x\dfrac{-3}{1-3x}


So you need to consider the inverse chain rule there


Ohhhh right!

So it would be simply 3[ln |2+x|](limits) - [ln |1-3x|](limits) instead of +?
Original post by MatthewP01
Ohhhh right!

So it would be simply 3[ln |2+x|](limits) - [ln |1-3x|](limits) instead of +?


you need to account for the 3 as well
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Original post by TenOfThem
you need to account for the 3 as well


So it would be [ln (1-3x) / 3]?
Original post by MatthewP01
So it would be [ln (1-3x) / 3]?


yes
Reply 10
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MatthewP01
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Original post by TenOfThem
yes


So with the limits I'm getting:

3ln2 - [(ln1 / 3) - (ln4 / 3)] sorry for bothering you again but how can you simply that to get it in the form pln2?
Reply 11
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MatthewP01
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Wait I got 3ln2 - 2/3ln2, so it would be 7/3ln2? I believe that is the right answer haha!
Original post by MatthewP01
So with the limits I'm getting:

3ln2 - [(ln1 / 3) - (ln4 / 3)] sorry for bothering you again but how can you simply that to get it in the form pln2?


The divide by 3 is not in the ln

3ln⁡2−13ln⁡1−13ln⁡43\ln 2 - \dfrac{1}{3}\ln 1 - \dfrac{1}{3}\ln 4

You need to realise that ln4 can be written in terms of ln2

Also do you know know what ln1 is
Reply 13
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MatthewP01
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Original post by TenOfThem
The divide by 3 is not in the ln

3ln⁡2−13ln⁡1−13ln⁡43\ln 2 - \dfrac{1}{3}\ln 1 - \dfrac{1}{3}\ln 4

You need to realise that ln4 can be written in terms of ln2

Also do you know know what ln1 is


Yeah ln1 is 0. I'm not sure, would it not just be ln2(sq)?
Original post by MatthewP01
Yeah ln1 is 0. I'm not sure, would it not just be ln2(sq)?


ln⁡4=ln⁡22\ln 4 = \ln 2^2 then what can you do?
Reply 15
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MatthewP01
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Original post by TenOfThem
ln⁡4=ln⁡22\ln 4 = \ln 2^2 then what can you do?


Just 2ln2?
Original post by MatthewP01
Just 2ln2?


yes
Reply 17
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MatthewP01
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Original post by TenOfThem
yes


So 3ln2 - (1/3)(2)ln2 would equal 7/3 ln2? think I finally got it!
Original post by MatthewP01
So 3ln2 - (1/3)(2)ln2 would equal 7/3 ln2? think I finally got it!


:biggrin:
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MatthewP01
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Original post by TenOfThem
:biggrin:


Thank you so much!!! C3 Maths is not my strong point! :colondollar:

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