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Thread: Improper Integral

  1. #1
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    Improper Integral

    Hi I need help on an improper integral.

    The integral from negative infinity to 3, where f(x)= 1/(x^2+9)

    I know I need to get it to the format, x^2+1 on the bottom, in order to get it to a tangent inverse; however, I am having trouble with my algebra to get to this point. Moreover we cannot use u-sub, because y' of x^2+9= 2x, and we cannot make an X; so this is our only option.
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  2. #2
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    Re: Improper Integral

    Divide both numerator and denominator by 9: \frac{1}{x^2+ 9}= \frac{\frac{1}{9}}{\left(\frac{x}{3}\right)^2+1} so that the substitution y= \frac{x}{3}, dy= \frac{1}{3}dx, and then 3dy= dx, gives \int_{-\infty}^3\frac{dx}{x^3+ 9}= 3\int_{-\infty}^1 \frac{dy}{y^2+ 1}.
    Thanks from math951
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