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Math Help - Integration

  1. #1
    Member kjchauhan's Avatar
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    Integration

    Please help me to solve the Integration:

    \int \frac{dx}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}

    Thanks in advance..
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  2. #2
    Senior Member DeMath's Avatar
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    Hint

    \displaystyle{\begin{aligned}\frac{1}{(x-1)^{3/4}(x+2)^{5/4}}&=\frac{1}{(x-1)^{3/4}(x+2)^{2-3/4}}=\frac{1}{(x-1)^{3/4}\dfrac{(x+2)^2}{(x+2)^{3/4}}}=\\[2pt]&=\frac{1}{{\left(\dfrac{x-1}{x+2}\right)\!}^{3/4}(x+2)^2}={\left(\frac{x-1}{x+2}\right)\!}^{-3/4}}\frac{1}{(x+2)^2}\end{aligned}}

    Now try to use this substitution \dfrac{x-1}{x+2}=u~\Rightarrow~\dfrac{dx}{(x+2)^2}=\dfrac{d  u}{3}
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  3. #3
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    Let t = \frac{x-1}{x+2} \Rightarrow \frac{dt}{dx} = \frac{3}{(x+2)^2} \Rightarrow dx = \frac{1}{3}(x+2)^2\;{dt}. Then we have:

    \displaystyle \begin{aligned} \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}\;{dx} & = \frac{1}{3}\int\frac{(x+2)^2}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}\;{dt} = \frac{1}{3}\int\frac{(x+2)^{\frac{3}{4}}}{(x-1)^{\frac{3}{4}}}\;{dt} \\& =  \frac{1}{3}\int\left(\frac{x+2}{x-1}\right)^{\frac{3}{4}}\;{dt} = \frac{1}{3}\int\frac{1}{t^{\frac{3}{4}}}\;{dt} = \frac{4}{3}t^{\frac{1}{4}}+k. \end{aligned}

    Thus \displaystyle \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dx} = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+k.

    EDIT: Beaten by four minutes. And I thought this solution was clever/unique!
    Last edited by TheCoffeeMachine; February 19th 2011 at 10:16 AM.
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  4. #4
    Senior Member DeMath's Avatar
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    Typos

    \displaystyle \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dx}  \ne \frac{1}{3}\int\frac{(x+2)^2}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dt} =\ldots
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  5. #5
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    Quote Originally Posted by DeMath View Post
    Typos
    4 instead of 3, right? If you're referring to that, then sorted, thanks.
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  6. #6
    Senior Member DeMath's Avatar
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    Sorry, at first, I did not understand the course of your solution.
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  7. #7
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    By parts, anyone? Do the addition/subtraction (how about 'numerator give-and-take'?!) thing -


    \displaystyle{\begin{aligned}\frac{1}{(x - 1)^{3/4}(x+2)^{5/4}} & = \frac{x + 2 - x - 1}{(x - 1)^{3/4}(x+2)^{5/4}} \\& = \frac{x + 2}{(x - 1)^{3/4}(x+2)^{5/4}} - \frac{x + 1}{(x - 1)^{3/4}(x+2)^{5/4}} \\& = \frac{1}{(x - 1)^{3/4}(x+2)^{1/4}} - \frac{x + 1}{(x - 1)^{3/4}(x+2)^{5/4}}\end{aligned}}<br />


    Then do integration by parts on the first half...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.



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  8. #8
    Member kjchauhan's Avatar
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    Thank you very much to all..
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