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Math Help - limit of a proper integral

  1. #1
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    limit of a proper integral

    How would you write:
    intergral from 2 to infinity of (x+3)/[(x^2-1)(x-1)]

    as the limit of a proper intergral, then evaluate the intergral.
    Then take the limit to find the improper definite intergral?

    The partial fraction is (1/2)ln(x+1) - (1/2)ln(x-1) - 2/(x-1)
    thanks
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  2. #2
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    Well you can set

    \int_2^\infty\frac{x+3}{(x^2-1)(x-1)}\,dx=\lim_{a\to\infty}\int_2^a\frac{x+3}{(x^2-1)(x-1)}\,dx

    Then solve.
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  3. #3
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    Yea i get that however i am unsure what to do when it comes out as
    [1/2ln(a+1)-2/(a-1)-1/2ln(a-1)]-[1/ln(3)-2]
    what do you do with the ln terms do you write ln(a+1/a-1)
    Then what do you do?
    Thanks
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  4. #4
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    \mathop {\lim }\limits_{a \to \infty } \int_2^a {\frac{{x + 3}}<br />
{{\left( {x^2 - 1} \right)\left( {x - 1} \right)}}\,dx} = \mathop {\lim }\limits_{a \to \infty } \left( {\left. { - \frac{1}<br />
{2}\ln \left| {x - 1} \right| + \frac{1}<br />
{2}\ln \left| {x + 1} \right| - \frac{2}<br />
{{x - 1}}} \right|_2^a } \right)

    So, when you're evaluating for a, turn back into \infty
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  5. #5
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    But then we get ln(infinity) which is infinty??
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  6. #6
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    Hello, Ben3535!

    What exactly is your question?
    It seems that you found the correct partial fractions and you integrated correctly.

    So what left?


    How would you write: intergral from 2 to infinity of (x+3)/[(x^2-1)(x-1)]

    as the limit of a proper intergral, then evaluate the intergral.
    Then take the limit to find the improper definite intergral?

    The partial fraction is (1/2)ln(x+1) - (1/2)ln(x-1) - 2/(x-1)

    Partial fractions: . \frac{x+3}{(x+1)(x-1)^2} \;=\;\frac{\frac{1}{2}}{x+1} + \frac{\text{-}\frac{1}{2}}{x-1} + \frac{2}{(x-1)^2}

    Integrate: . \frac{1}{2}\int\frac{dx}{x+1} - \frac{1}{2}\int\frac{dx}{x-1} + 2\int(x-1)^{-2}dx

    . . =\;\frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x-1) - \frac{2}{x-1} \;=\;\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right) - \frac{2}{x=1}


    We will evaluate from 2 to \infty

    . . \lim_{b\to\infty}\bigg[\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right) - \frac{2}{x-1}\bigg]^b_2 \;= \;\lim_{b\to\infty}\bigg[\left(\frac{1}{2}\ln\frac{b+1}{b-1} - \frac{2}{b-1}\right) - \left(\frac{1}{2}\ln\left(\frac{3}{1}\right) - \frac{2}{1}\right)\bigg]


    In the first fraction, divide top and bottom by b.

    . . \lim_{b\to\infty}\bigg[\frac{1}{2}\ln\left(\frac{1 + \frac{1}{b}}{1 - \frac{1}{b}}\right) - \frac{2}{b-1} - \frac{1}{2}\ln(3) + 2\bigg]

    . . =\;\frac{1}{2}\ln\left(\frac{1+0}{1-0}\right) - 0 - \frac{1}{2}\ln(3) + 2

    . . = \;\frac{1}{2}\ln(1) - \frac{1}{2}\ln(3) + 2 \;=\;\boxed{2 - \frac{1}{2}\ln(3)}

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