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Math Help - Indefinite Integral

  1. #1
    Member sbhatnagar's Avatar
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    Indefinite Integral

    What is the value of \int \frac{x+2}{x^2+x+1}dx?

    Can it be solved? If yes, what is the answer?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Indefinite Integral

    Yes it can be solved, but you have to rewrite some things. We want to recognize the first derivative of the denominator which is 2x+1 in the numerator, so:
    \int \frac{x+2}{x^2+x+1}dx =\frac{1}{2} \int \frac{2x+4}{x^2+x+1}dx=\frac{1}{2}\int \frac{2x+1+3}{x^2+x+1}dx =\frac{1}{2}\int \frac{2x+1}{x^2+x+1}dx +\frac{3}{2}\int \frac{dx}{x^2+x+1}

    For the second integral use completing the square in the denominator.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Indefinite Integral

    Have I done correct?

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

    =\frac{1}{2} \int \frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int \frac{1}{x^2+x+1}dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{x^2+x+1}dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{(\frac{2x+1}{2})^2+(\frac{\sqrt{3}}{2})^2  }dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\sqrt{3}\tan^{-1}(\frac{2x+1}{\sqrt{3}})
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  4. #4
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    Re: Indefinite Integral

    Quote Originally Posted by sbhatnagar View Post
    Have I done correct?

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

    =\frac{1}{2} \int \frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int \frac{1}{x^2+x+1}dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{x^2+x+1}dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{(\frac{2x+1}{2})^2+(\frac{\sqrt{3}}{2})^2  }dx

    =\frac{1}{2} \ln{(x^2+x+1)}+\sqrt{3}\tan^{-1}(\frac{2x+1}{\sqrt{3}})
    Why didn't you do as instructed and complete the square in the denominator of the second integral?
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