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Math Help - Integration

  1. #1
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    Integration

    I can't seem to get the answer to

    2xy^3/(x^2y^2+1)
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  2. #2
    MHF Contributor harish21's Avatar
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    Please mention if it is

    \displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dx

    or

    \displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dy

    and show your attempt.
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  3. #3
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    Sorry it is dy

    I tried a substitution and it got very messy, ended up with y= sqrt( u-1/x). Hence I think I have taken the completely wrong route
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  4. #4
    MHF Contributor harish21's Avatar
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    \displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dy

    = \displaystyle 2x \; \int \dfrac{y^3}{x^2y^2+1} \;dy

    Use Partial fractions:

     \dfrac{y^3}{x^2y^2+1} = \dfrac{y}{x^2} - \dfrac{y}{x^2 (x^2 y^2+1)} dy
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  5. #5
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    Or you may make the substitution x^{2}y^{2} = t,

    so that dt = 2x^{2}ydy

    Therefore dy = dt/(2x^{2}y)

    and y^{2} = t/x^{2}
    Therefore the integrand now becomes, \int(1/x^{3})(t/(t+1))\;dt

    This can be evaluated using standard forms.
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  6. #6
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    Thank you

    I can integrate y/x^2 dy to get y^2/2x^2

    Do I tackle the second integral by substitution?

    I got du=2x^4y dy and my result for y again seems wrong as I got square root of -1/x^2 which you can't have
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  7. #7
    MHF Contributor harish21's Avatar
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    \displaystyle \int \dfrac{y^3}{x^2y^2+1} =\displaystyle  \int 2x \dfrac{y}{x^2}dy - \displaystyle  \int 2x \dfrac{y}{x^2 (x^2 y^2+1)} dy

    =\displaystyle  \int 2 \dfrac{y}{x} \;dy -\displaystyle  \int 2 \dfrac{y}{x(x^2 y^2+1)} dy

    for the second one, use substution.

    u=x^2y^2+1 \; \therefore du = x^2(2y)\;dy and finish..
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