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integrating using partial fraction method Watch

    • Thread Starter
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    Hey guys, I was hoping someone could check over my assignment answer, I havent had too much experience with this type of question... thanks


    ∫(x+6)/x²-2x-8

    using partial fractions approach we get:

    x²-2x-8=(x+2)+(x+4)

    so:

    (x+6)/x²-2x-8=(x+6)/(x+2)+(x+4)

    =(A/(x+2))+(B/(x-4))

    multiplying through by (x+2)+(x-4) gives:

    x+6=A(x-4)+B(x+2)

    let x=4

    (4)+6=B(4+2)

    B=5/3

    let x=-2

    (-2)+6=A(-2-4)

    A=-2/3

    therefore:

    ∫(x+6)/x²-2x-8dx=∫((-2/3)/x-4)+((5/3)/x+2)dx

    =∫((-2/3)(1/x-4)+(5/3)(1/x+2))dx

    =ln[(x+2)5/3]-ln[(x-4)2/3]+c

    =ln[(x+2)5/3]/[(x-4)2/3]+c
    • Community Assistant
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    I don't think we ought to help with your assignment but you can use Wolfram Alpha to evaluate integrals.

    Right at the start of your working you added two linear terms to make a quadratic. I didn't read further.
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    (Original post by Bourbon daddy)


    x²-2x-8=(x+2)+(x+4)
    You need to check this!
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    Your working is correct except for that graphical error where you put plus signs instead of multiplication signs >.>
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    thanks for pointing that site out. I have actually been told about it by my lecturer but I have refrained from using it as I didnt want the temptation of using it to help too much. Having said that, I did just have a quick look and it looks like I have mixed up my terms somewhere along the line...

    one last thing, am I right in thinking, a law of indices suggests

    5/3log(4-x)

    can also be written:

    log(4-x)^5/3
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    (Original post by Bourbon daddy)
    thanks for pointing that site out. I have actually been told about it by my lecturer but I have refrained from using it as I didnt want the temptation of using it to help too much. Having said that, I did just have a quick look and it looks like I have mixed up my terms somewhere along the line...

    one last thing, am I right in thinking, a law of indices suggests

    5/3log(4-x)

    can also be written:

    log(4-x)^5/3
    Yes.
    • Thread Starter
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    Hey Davros... you are correct, sorry it was a typo, it was meant to read x-4... I am trying to get some sneaky study time in whilst looking after my one year old and he was getting a bit grumpy on my lap when I was typing it up... he is asleep now... thanks for all the posts, I appreciate all of your help
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    Yes, that is true, and your answer is correct according to wolfram alpha.
 
 
 
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