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

    Zacken
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    (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.
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    (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.
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    (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
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    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
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    I have tried doing this
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    (Original post by loooolo12345)
    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
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    I have tried doing this
    Nah, long division first!
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    (Original post by loooolo12345)
    Name:  Photo on 07-03-2016 at 19.09 #2.jpg
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    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.
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    (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?
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    (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.
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    (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
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    (Original post by loooolo12345)
    Ok awesome, I can do the rest now Thank you very much guys
    Glad to have helped!
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    If the degree is the same there should also be constant, so your approach for partial fractions wasn't correct


    Posted from TSR Mobile
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    (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!
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    (Original post by Bath_Student)
    It was all correct. Have another check!
    In their initial working?


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    (Original post by drandy76)
    In their initial working?


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    Indeed, I see. The working was redundant though!
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    (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


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    (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


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    They're equivalent.
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    (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
 
 
 

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