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Math Help - Improper Integrals

  1. #1
    Senior Member mollymcf2009's Avatar
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    Improper Integrals

    \int\limits^{\infty}_{0} 4 \frac{e^x}{e^{2x}+3} dx

    Just want to make sure I am starting this correctly.

    I rewrote as:

    \int\limits^{\infty}_{0} 4 \frac{e^x}{(e^x)^2+3} dx

    Then did a u-substitution with e^x and got:

    4 \int \frac{u}{u^2+3} du

    Then did another sub with v= u^2 + 3 and got:

    2 \int \frac{1}{v} dv

    Integrated and got:

    2 [ln|u^2+3|] = 2[ln|e^{2x}+3|]

    Is this how I should be setting this up before I do my limit for my improper integral?

    Thanks!
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  2. #2
    o_O
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    \lim_{t \to \infty} \int_0^{t} \frac{e^x}{(e^x)^2 + 3}\  dx

    Actually: {\color{red}u = e^x} \ \Rightarrow \ {\color{blue}du = e^x dx}

    (Don't forget to change the limits!)

    So: \lim_{t \to \infty} \int_0^{t} \frac{{\color{blue}e^xdx}}{({\color{red}e^x})^2 + 3} =  \lim_{t \to \infty}\int_{1}^{t} \frac{{\color{blue}du}}{{\color{red}u}^2 + 3}

    And this should remind you of the integral: \int \frac{dx}{x^2 + a^2} = \tfrac{1}{a} \arctan \left(\tfrac{x}{a}\right) + C
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Thanks!!

    Here is what I got:

    \lim_{t \to \infty} \frac{4}{\sqrt3} [tan^{-1} \frac{e^t}{\sqrt3} - tan^{-1} \frac{1}{\sqrt3}]|^{t}_{0}

    = \frac{4}{\sqrt3} [\frac{\pi}{2} - \frac{\pi}{6}]

    Look okay?
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