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    How do I integrate x^2/4-x?

    Thank you!
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    (Original post by Hlrk)
    How do I integrate x^2/4-x?

    Thank you!
    Carry out some sort of division and integrate term by term.
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    How would I go about that?
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    (Original post by Hlrk)
    How would I go about that?
    Using your preferred method of polynomial division.
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    If you can't remember how to divide polynomials you could note that:

    \displaystyle \frac{x^2}{4-x}=\frac{x^2-16+16}{4-x}=\frac{(x+4)(x-4)+16}{4-x}=-(x+4)+\frac{16}{4-x}
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    (Original post by Hlrk)
    How do I integrate x^2/4-x?

    Thank you!
    An alternative method would be to substitute u=4-x
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    (Original post by Mr M)
    If you can't remember how to divide polynomials you could note that:

    \displaystyle \frac{x^2}{4-x}=\frac{x^2-16+16}{4-x}=\frac{(x+4)(x-4)+16}{4-x}=-(x+4)+\frac{16}{4-x}
    Huh...I have never approached it like that!
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    (Original post by keromedic)
    Huh...I have never approached it like that!
    It's just another way of approaching the problem of rewriting the integrand in a simpler way - polynomial long division will always work, but if you can spot a trick like this it always helps!
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    (Original post by davros)
    It's just another way of approaching the problem of rewriting the integrand in a simpler way - polynomial long division will always work, but if you can spot a trick like this it always helps!
    Thanks
    I get how this works. I hate long division so kind of do what MrM did but rather unintuively Through use of the remainder theorem.
    Think it goes something like
    \dfrac{P(x)}{Q(x)}=R(x)+\dfrac{S  (x)}{Q(x)}..
 
 
 
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