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  1. #1
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    Need Help Starting Problem

    Evaluate the integral by making a substitution that converts the integrand to a rational function
    $\displaystyle \int\ \frac{e^t}{e^{2t}-4}dx$
    can i do the problem this way
    $\displaystyle \int\ \frac{e^t}{e^{t^2}-4}dx$
    then doin a u sub with $\displaystyle u=e^t$
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  2. #2
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    Quote Originally Posted by vinson24 View Post
    Evaluate the integral by making a substitution that converts the integrand to a rational function
    $\displaystyle \int\ \frac{e^t}{e^{2t}-4}dx$
    can i do the problem this way
    $\displaystyle \int\ \frac{e^t}{e^{t^2}-4}dx$
    then doin a u sub with $\displaystyle u=e^t$
    $\displaystyle \int {\frac{{du}}
    {{u^2 - 4}} = \frac{1}
    {4}\int {\left( {\frac{1}
    {{u - 2}} - \frac{1}
    {{u + 2}}} \right)} du} $
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  3. #3
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    Quote Originally Posted by Plato View Post
    $\displaystyle \int {\frac{{du}}
    {{u^2 - 4}} = \frac{1}
    {4}\int {\left( {\frac{1}
    {{u - 2}} - \frac{1}
    {{u + 2}}} \right)} du} $
    How did you get the 1/4?
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  4. #4
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    $\displaystyle \frac{1}
    {{u - 2}} - \frac{1}
    {{u + 2}} = \frac{4}
    {{u^2 - 4}}$
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by vinson24 View Post
    Evaluate the integral by making a substitution that converts the integrand to a rational function
    $\displaystyle \int\ \frac{e^t}{e^{2t}-4}dx$
    can i do the problem this way
    $\displaystyle \int\ \frac{e^t}{e^{t^2}-4}dx$
    then doin a u sub with $\displaystyle u=e^t$
    $\displaystyle \int {\frac{e^y}{-4+e^{2t}}} dt $

    substitute $\displaystyle u = e^t $ and $\displaystyle du = e^{t} dt$ :

    = $\displaystyle \int {\frac{1}{u^{2}-4}} du$


    = $\displaystyle \frac{-1}{2} tanh^{-1}\frac{u}{2}+C$

    Substitute back for $\displaystyle u = e^t$:

    = $\displaystyle \frac{-1}{2}tanh^{-1}\frac{e^t}{2}+C$

    Which is equivalent for t values to:

    = $\displaystyle \frac{1}{4} log(e^{t}-2)-\frac{1}{4} log(e^{t}+2)+C $
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