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Math Help - Improper Integrals 2

  1. #1
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    Improper Integrals 2

    evaluate the integral or state that it diverges:

    integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx
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  2. #2
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    Partial fractions should do the job:

     \int_0^1 \frac{x+1}{x^2+2x}dx = \int_0^1 \frac{x+1}{x(x+2)}dx = \int_0^1 \left[\frac{A}{x} + \frac{B}{x+2}\right]dx

     A(x+2) + Bx = x+1

    set x = -2 to get B; set x = 0 to get A.
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  3. #3
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    the problem has been misread, the integrand is actually \frac{x+1}{\sqrt{x^2+2x}}.
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  4. #4
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    Quote Originally Posted by smartartbug View Post
    evaluate the integral or state that it diverges:

    integral from 0 to 1 of ((x+1)/sqrt((x^2)+2x))dx

    \int_{0}^1\frac{(x+1)}{\sqrt{x^2+2x}}dx

    let u=x^2+2x so du=2(x+1)dx

    \int_{0}^3\frac{(x+1)}{2(x+1)\sqrt{u}}du

    \frac{1}{2}\int_{0}^3u^{-\frac{1}{2}}du
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  5. #5
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    just to mention that the integral converges by limit comparison test with \int_0^1\frac{dx}{\sqrt x}.
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