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Math Help - separable equation

  1. #1
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    separable equation

    (x+1)\frac{dy}{dx}=x(y^2+1)

    i basically get how to do the separable equations, i change it to:

    \int{\frac{dy}{(y^2+1)}}=\int{\frac{xdx}{(x+1)}}

    and then i integrate both sides, i'm having trouble integrating the right side, i tried using integration by parts but it doesn't come out how the book has it. i think maybe there's an easier way but i can't remember. can someone help me with this?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Let u = x+1

    then the integrand becomes (u-1)/u = 1 - 1/u

    Or use long division to obtain 1 - 1/(x+1) directly

    at any rate you get x - ln(x+1) + c aftger integrating
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    Let u = x+1

    then the integrand becomes (u-1)/u = 1 - 1/u

    Or use long division to obtain 1 - 1/(x+1) directly

    at any rate you get x - ln(x+1) + c aftger integrating
    No need for long division!

    \frac{x}{x+1} = \frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}
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