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Need help with 2 integration problems... watch

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    • Thread Starter
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    Hey guys, Im sorry, I just dont know where to start with these 2 questions...
    Any help would be greatly appreciated.
    Thanks in advance

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    For the first one, differentiate the quadratic on the bottom (not the cube bit, just the quadratic) and see if this helps.
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    First one, do substitution. Second one, divide top by bottom and then integrate.

    edit: Divide as in polynomial division
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    First one, make a substitution for the bit inside the bracket on the denominator.

    Second one, express as a quotient and two partial fractions.
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    for a) make a substitution, then see how the derivative of the substitution relates to the numerator of the fraction

    ah... beaten
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    use the substitution u=x^2-4x +3
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    (Original post by Mr M)
    First one, make a substitution for the bit inside the bracket on the denominator.

    Second one, express as a quotient and two partial fractions.
    Would it not be easier to do polynomial division on the second one?
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    The first one is very simple because of the derivative of the bottom being a multiple of the top, however the second one requires substitution.
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    (Original post by PeeWeeDan)
    The first one is very simple because of the derivative of the bottom being a multiple of the top, however the second one requires substitution.
    How's that?
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    (Original post by D-Day)
    How's that?
    No idea I guessed just from glancing at it. Do I need to give it another looking at?
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    (Original post by D-Day)
    Would it not be easier to do polynomial division on the second one?
    I don't think so. The remainder would still need some work.

    Expressing it in this form sorts it all out in one go.

    \frac{3x^2+x-5}{x^2-4x}=A+\frac{B}{x}+\frac{C}{x-4}.
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    (Original post by PeeWeeDan)
    No idea I guessed just from glancing at it. Do I need to give it another looking at?
    Because top and bottom have the same power, substitution isn't the best method.
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    Too much talk of substitutions for the first one, surely it's quicker to notice
     \frac {d}{dx} (x^2 - 4x + 3)^-^2 = -2(2x-4)(x^2 - 4x + 3)^-^3
    Which is a multiple of the integral you want to do.
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    (Original post by D-Day)
    Because top and bottom have the same power, substitution isn't the best method.
    Fair enough.
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    (Original post by benwellsday)
    Too much talk of substitutions for the first one, surely it's quicker to notice
     \frac {d}{dx} (x^2 - 4x + 3)^-^2 = -2(2x-4)(x^2 - 4x + 3)^-^3
    Which is a multiple of the integral you want to do.
    ...that's why you do substitution...
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    What? No, from there I would just say
     -\frac {1}{4} \frac{d}{dx} (x^2 - 4x + 3)^-^2 =  \frac {x-2}{(x^2 - 4x + 3)^3}
    Then you have the answer without integration by substitution. Specifically
     -\frac{1}{4} \int \frac {d}{dx} (x^2 - 4x + 3)^-^2 dx = -\frac{1}{4} (x^2 - 4x + 3)^-^2 = \int \frac {x-2}{(x^2 - 4x + 3)^3} dx
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    (Original post by Mr M)
    I don't think so. The remainder would still need some work.

    Expressing it in this form sorts it all out in one go.

    \frac{3x^2+x-5}{x^2-4x}=A+\frac{B}{x}+\frac{C}{x-4}.
    How did you get that form? Where do I go from there? How do I find A, B and C? Thanks
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    any help with the 2nd question guys? thanks!
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    Express it in partial fractions. It comes out as the form you just quoted. To find A, B and C you can make it back into a single fraction (by multiplying by the denominators) and then make sure the x^2, x, and constant terms match up for the numerator. The integration is then pretty easy.
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    u cant express 2nd question as a partial fraction becase its a "top havic" which means that the powers of x are same so u cant do that.however i dont know how should u do dat question. I am surethat it cant b expree as partial fraction.
 
 
 
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