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Thread: Rational Function Integration

  1. #1
    Nov 2008

    Rational Function Integration

    Find the indefinite integral of the function:
    Attached Thumbnails Attached Thumbnails Rational Function Integration-problem-4.jpg  
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  2. #2
    MHF Contributor
    Oct 2005
    Doing this on Mathematica shows the answer has three separate terms which makes me think that partial fractions will be involved. I would try factoring the denominator and see where that leads.
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  3. #3
    Super Member redsoxfan325's Avatar
    Feb 2009
    Swampscott, MA
    Start with partial fractions.

    The denominator factors to: (x-1)(x^2-2x+10).

    Thus you have \frac{2x-29}{(x-1)(x^2-2x+10)}.

    Partial fractions: \frac{2x-29}{(x-1)(x^2-2x+10)} = \frac{A}{x-1}+\frac{Bx+C}{x^2-2x+10}

    Eliminating the denominators yields 2x-29 = A(x^2-2x+10) + (Bx+C)(x-1).

    Multiplying this out gives 2x-29 = (A+B)x^2 + (-2A-B+C)x + (10A-C).

    Set A+B = 0, -2A-B+C=2, and 10A-C=-29.

    This makes A=-3, B=3, and C=-1.

    Thus you integral becomes -3\int \frac{1}{x-1}\,dx + \int \frac{3x-1}{x^2-2x+10}\,dx. (I completed the square in the denominator of the second integral.)

    The first integral clearly equals -3ln(x-1).

    From now on, I'll be dealing only with the second integral.

    Rewrite it like this: \int \frac{3x-1-2+2}{x^2-2x+10}\,dx.

    Now, break it up into two integrals: \int\frac{3x-3}{x^2-2x+10}\,dx + \int\frac{2}{x^2-2x+10}\,dx.

    In the first integral, let u=x^2-2x+10. Thus du=2x-2\,dx. Now that integral becomes \frac{3}{2}\int\frac{\,du}{u} = \frac{3}{2}ln(u) = \frac{3}{2}ln(x^2+2x+10).

    On the second integral, factor out the 2 and complete the square in the denominator to get 2\int\frac{1}{(x-1)^2+9}\,dx.

    Then, factor a 9 out of the denominator to get: \frac{2}{9}\int\frac{1}{\frac{(x-1)^2}{9}+1}\,dx.

    Rewrite the denominator: \frac{2}{9}\int\frac{1}{(\frac{x-1}{3})^2+1}\,dx.

    Let u = \frac{x-1}{3} and du = \frac{1}{3}\,dx.

    Thus your integral becomes: \frac{2}{9}\int\frac{3\,du}{u^2+1}.

    Factor out the 3 and you have \frac{2}{3}\int\frac{\,du}{u^2+1} = \frac{2}{3}arctan(u) = \frac{2}{3}arctan(\frac{x-1}{3}).

    Adding all the parts together gives you: -3ln(x-1) + \frac{3}{2}ln(x^2+2x+10) + \frac{2}{3}arctan(\frac{x-1}{3}).

    Thus: \int\frac{2x-29}{x^3-3x^2+12x-10}\,dx = -3ln(x-1) + \frac{3}{2}ln(x^2+2x+10) + \frac{2}{3}arctan(\frac{x-1}{3}).
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