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How to integrate (2x)(3-x)^-1) C4 integration Watch

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    (2x)(3-x)^-1
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    Moved to maths.

    What have you tried so far? What's the full question?
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    Try partial fractions.
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    Make it ((3-X)^-1)(2x) and integrate by parts? That's how I'd do it I think lol?

    Dont think its a log so
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    (Original post by RonnieRJ)
    Make it ((3-X)^-1)(2x) and integrate by parts? That's how I'd do it I think lol?

    Dont think its a log so
    You've just re-written the integrand with the brackets in opposite places, and integration by parts is not a viable approach here.

    There is a logarithmic term in the evaluation...
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    (Original post by family'sfailure)
    (2x)(3-x)^-1
    \displaystyle - 2\frac{x}{x-3} = -2 \left(\frac{x-3 + 3}{x-3}\right) = -2 \left(1 + \frac{3}{x-3}\right)
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    (2x)(3-x)^-1

    (Original post by alow)
    Try partial fractions.
    If 2x/(3-x), how can that be decomposed?

    I would write \dfrac{2x}{3-x} as \dfrac{-(6-2x)}{3-x}+\dfrac{6}{3-x}
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    (Original post by Zacken)
    You've just re-written the integrand with the brackets in opposite places, and integration by parts is not a viable approach here.

    There is a logarithmic term in the evaluation...
    Yeah exactly, so that you don't have to try and raise the power of the bracket by one

    And oh whoops didn't look at it that way
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    (Original post by RonnieRJ)
    Yeah exactly, so that you don't have to try and raise the power of the bracket by one
    What...?
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    (Original post by Zacken)
    What...?
    Dw it works in my head lmao I appreciate you'd do it differently
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    (Original post by Kvothe the arcane)
    (2x)(3-x)^-1



    If 2x/(3-x), how can that be decomposed?

    I would write \dfrac{2x}{3-x} as \dfrac{-(6-2x)}{3-x}+\dfrac{6}{3-x}
    \dfrac{6}{3-x} -2
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    (Original post by RonnieRJ)
    Dw it works in my head lmao I appreciate you'd do it differently
    No
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    (Original post by Kvothe the arcane)
    No
    No?
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    (Original post by alow)
    \dfrac{6}{3-x} -2
    Yes . Sorry! I phrased my question wrongly.

    Is the above also considered partial fractions? I thought that partial fraction decomposition was concerned with the decomposition of fractions with algebraic denominators of order 2 or above.
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    (Original post by RonnieRJ)
    No?
    It is not that Zackan is saying that that you don't know how to integrate by parts.

    He is saying that it's not the correct technique to use here.
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    (Original post by Kvothe the arcane)
    Yes . Sorry! I phrased my question wrongly.

    Is the above also considered partial fractions? I thought that partial fraction decomposition was concerned with the decomposition of fractions with algebraic denominators with order 2 or above.
    I'd still call it that because you use the same method just with a constant factor (i.e. a factor with denominator 1).
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    (Original post by Kvothe the arcane)
    It is not that Zackan is saying that that you don't know how to integrate by parts.

    He is saying that it's not the correct technique to use here.
    That's what I said lmao? I said I now appreciate you'd do it a different way lol
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    (Original post by Zacken)
    \displaystyle - 2\frac{x}{x-3} = -2 \left(\frac{x-3 + 3}{x-3}\right) = -2 \left(1 + \frac{3}{x-3}\right)
    You really do love this trick
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    (Original post by kingaaran)
    You really do love this trick
    It's a nice one
 
 
 
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