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Integral with sq. root in it again

  1. Nov 27, 2007 #1

    rock.freak667

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    Integral with sq. root in it....again

    1. The problem statement, all variables and given/known data
    Find..
    [tex]\int x^\frac{3}{2}\sqrt{1+x} dx[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Well I used the fact that:

    [tex]\sqrt{1+x}=\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^n}{(1-2n)(n!)^24^n}[/tex]

    and well I just multiplied by [itex]x^\frac{3}{2}[/itex]

    so I integrated:
    [tex]\int \sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{3}{2}}{(1-2n)(n!)^24^n}[/tex]

    and got [tex]\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{5}{2}}{(1-2n)(n!)^24^n\frac{5}{2}}[/tex]

    [itex]\frac{2x^\frac{5}{2}}{5}\sqrt{1+x}[/itex] which is wrong because if i differentiate it I get an extra term in it
     
  2. jcsd
  3. Nov 28, 2007 #2

    Gib Z

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    That Series is only valid for |x| < 1. I would let [itex]x= \sinh^2 u[/itex]
     
  4. Nov 28, 2007 #3

    rock.freak667

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    that hyperbolic substitution throws me off as it kinda made it harder for me
     
  5. Nov 28, 2007 #4

    Gib Z

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    With that substitution I get: [tex]2\int \sinh^4 u \cosh^2 u du[/tex], which I believe is possible through methods similar to its circular trigonometric counterpart.

    EDIT: I've just done it, its not as easy as I thought but it is possible. Express all the squares in terms of double angle formula. You should get an answer with an x term, and the hyperbolic sines of 2x, 4x and 6x.
     
    Last edited: Nov 28, 2007
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