Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    Please help me with this integral:
    (2x^2-x)/((x-2)(x+1))
    I have tried partial fractions but this hasn't worked

    Zacken
    Offline

    3
    ReputationRep:
    (Original post by loooolo12345)
    Please help me with this integral:
    (2x^2-x)/((x-2)(x+1))
    I have tried partial fractions but this hasn't worked
    Expand denominator and use long division would be my first guess.
    Online

    22
    ReputationRep:
    (Original post by loooolo12345)
    Please help me with this integral:
    (2x^2-x)/((x-2)(x+1))
    I have tried partial fractions but this hasn't worked

    Zacken
    You need to do long division first because the numerator is of the same degree as the denominator: see this. Is that helpful? If not, come back after you've watched it and show me your working out.
    Offline

    18
    ReputationRep:
    (Original post by loooolo12345)
    Please help me with this integral:
    (2x^2-x)/((x-2)(x+1))
    I have tried partial fractions but this hasn't worked

    Zacken
    its improper, did you use long division first?

    Edit: maths forum is 2 quick
    Spoiler:
    Show
    i know im not zacken but ill have to do
    • Thread Starter
    Offline

    1
    ReputationRep:
    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
Views: 75
Size:  39.7 KB

    I have tried doing this
    Offline

    3
    ReputationRep:
    (Original post by loooolo12345)
    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
Views: 75
Size:  39.7 KB

    I have tried doing this
    Nah, long division first!
    Online

    22
    ReputationRep:
    (Original post by loooolo12345)
    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
Views: 75
Size:  39.7 KB

    I have tried doing this
    You can't do partial fractions unless the degree of the denominator exceed the degree of the numerator. See the video I linked.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Zacken)
    You can't do partial fractions unless the degree of the denominator exceed the degree of the numerator. See the video I linked.
    ok, thanks, So I have watched the video and now have that the fraction is equal to:

    2+((x+4))/((x+1)(x-2))

    Is this right?
    Online

    22
    ReputationRep:
    (Original post by loooolo12345)
    ok, thanks, So I have watched the video and now have that the fraction is equal to:

    2+((x+4))/((x+1)(x-2))

    Is this right?
    That's very much correct. Now you can do partial fractions on the second term because the degree of the bottom exceeds the degree of the top.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Zacken)
    That's very much correct. Now you can do partial fractions on the second term because the degree of the bottom exceeds the degree of the top.
    Ok awesome, I can do the rest now Thank you very much guys
    Online

    22
    ReputationRep:
    (Original post by loooolo12345)
    Ok awesome, I can do the rest now Thank you very much guys
    Glad to have helped!
    Offline

    12
    ReputationRep:
    If the degree is the same there should also be constant, so your approach for partial fractions wasn't correct


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by drandy76)
    If the degree is the same there should also be constant, so your approach for partial fractions wasn't correct


    Posted from TSR Mobile
    It was all correct. Have another check!
    Offline

    12
    ReputationRep:
    (Original post by Bath_Student)
    It was all correct. Have another check!
    In their initial working?


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by drandy76)
    In their initial working?


    Posted from TSR Mobile
    Indeed, I see. The working was redundant though!
    Offline

    12
    ReputationRep:
    (Original post by Bath_Student)
    Indeed, I see. The working was redundant though!
    Adding the constant instead of using long division is how I was taught, didn't even realise that was a method of doing it until seeing this thread


    Posted from TSR Mobile
    Online

    22
    ReputationRep:
    (Original post by drandy76)
    Adding the constant instead of using long division is how I was taught, didn't even realise that was a method of doing it until seeing this thread


    Posted from TSR Mobile
    They're equivalent.
    Offline

    12
    ReputationRep:
    (Original post by Zacken)
    They're equivalent.
    the second i learned how to equate coefficients i refused to do long division ever again, no ragrets
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.