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Math Help - Integral of Rational Function

  1. #1
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    Integral of Rational Function

    I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

    \int{\frac{1}{x^2 + x + 1}}

    But this can't be factored and completing the square gives (x+\frac{1}{2})^2+\frac{3}{4}

    Which doesn't look any easier to integrate.

    Could someone point me in the right direction?

    Thanks
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  2. #2
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    Quote Originally Posted by centenial View Post
    I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

    \int{\frac{1}{x^2 + x + 1}}

    But this can't be factored and completing the square gives (x+\frac{1}{2})^2+\frac{3}{4}

    Which doesn't look any easier to integrate.

    Could someone point me in the right direction?

    Thanks
    You missed a dx.

    Completing the square is a good start, move on by substituting u=x+1/2.
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  3. #3
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    Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.

    So...

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

    <br />
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}<br />

    <br />
\int{\frac{1}{u^2 + \frac{3}{4}}du}<br />

    <br />
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}<br />

    <br />
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C<br />

    <br />
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C<br />

    Is that right?
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  4. #4
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    Quote Originally Posted by centenial View Post
    Oh... complete the square and then u substitution. That seems so obvious now that I can't believe I didn't see it.

    So...

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

    <br />
\int{\frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}}dx}<br />

    <br />
\int{\frac{1}{u^2 + \frac{3}{4}}du}<br />

    <br />
\int{\frac{1}{u^2 + \sqrt{\frac{3}{4}}^2}du}<br />

    <br />
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2u}{\sqrt{3}}} + C<br />

    <br />
\frac{2}{\sqrt{3}} \times \tan^{-1}{\frac{2x + 1}{\sqrt{3}}} + C<br />

    Is that right?
    Looks fine to me.
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  5. #5
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    Quote Originally Posted by centenial View Post
    I got this question on an exam this morning and I didn't know how to answer it. I've been working on it the past hour trying to figure it out. Usually with rational functions, I factor to get it in a from that I can use partial fractions.

    \int{\frac{1}{x^2 + x + 1}}

    But this can't be factored and completing the square gives (x+\frac{1}{2})^2+\frac{3}{4}

    Which doesn't look any easier to integrate.

    Could someone point me in the right direction?

    Thanks
    If you don't want to have to resort to two substitutions, substitute x + 1 = \frac{\sqrt{3}}{2}\tan{\theta} so that dx = \frac{\sqrt{3}}{2}\sec^2{\theta}\,d\theta.


    Then \int{\frac{dx}{x^2 + x + 1}} = \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the  ta}{\left(\frac{\sqrt{3}}{2}\tan{\theta}\right)^2 + \frac{3}{4}}}

     = \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the  ta}{\frac{3}{4}(\tan^2{\theta} + 1)}}

     = \int{\frac{\frac{\sqrt{3}}{2}\sec^2{\theta}\,d\the  ta}{\frac{3}{4}\sec^2{\theta}}}

     = \int{\frac{\frac{\sqrt{3}}{2}}{\frac{3}{4}}\,d\the  ta}

     = \int{\frac{2\sqrt{3}}{3}\,d\theta}

     = \frac{2\sqrt{3}\,\theta}{3} + C

     = \frac{2\sqrt{3}\arctan{\left[\frac{2\sqrt{3}}{3}(x+1)\right]}}{3} + C.
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