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Math Help - Completing the square and integration

  1. #1
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    Completing the square and integration

    <br />
\int \frac{5}{x^2+x-1} dx<br /> <br />

    I know that  \frac{\sqrt{5}}{2} sec\theta = x + 1/2

    but its so hard for me to get a clear answer.

    Can someone spare some time for me and show me how this is done?

    Thank You so much
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  2. #2
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     x^2+x-1= \left( x+\frac{1}{2}\right)^2-\frac{5}{4}

    now make the substitution.
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  3. #3
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    Quote Originally Posted by pickslides View Post
     x^2+x-1= \left( x+\frac{1}{2}\right)^2-\frac{5}{4}

    now make the substitution.
    i know but what im getting is  2\sqrt{5}\int csc\theta d\theta

    i dont know what to do from there to get a clear and reduced answered
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  4. #4
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    multiply by 1 = \frac{\csc \theta + \cot \theta}{\csc \theta + \cot \theta}

    then you have 2\sqrt{5}\int \frac{\csc^2 \theta + \csc \theta \cot \theta}{\csc \theta + \cot \theta} d \theta

    which is of the form \int \frac {du}{u}.
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    multiply by 1 = \frac{\csc \theta + \cot \theta}{\csc \theta + \cot \theta}

    then you have 2\sqrt{5}\int \frac{\csc^2 \theta + \csc \theta \cot \theta}{\csc \theta + \cot \theta} d \theta

    which is of the form \int \frac {du}{u}.
    thank you but i know the integral of that but what I'm looking for the answer and how gotten to it.

     -\ln \mid \csc\theta + \cot\theta \mid + c
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