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Math Help - Integration question

  1. #1
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    Integration question

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  2. #2
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    Note \frac{4x^2 + 3x + 1}{x(2x + 1)} = \frac{1}{x} - \frac{1}{2x + 1} + 2.
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  3. #3
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    How do you know to break it up like this?
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  4. #4
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    For integrands involving improper polynomial fractions, it's a common technique to rewrite the integrand as a proper polynomial fraction. Improper polynomial fractions are fractions in which the highest degree in the numerator is greater than or equal to the highest degree in the denominator.
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  5. #5
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    Quote Originally Posted by fterh View Post
    Judging by the hint, I think here is the way they want you to do it:

    Split the integral into two:

    \displaystyle \int_{1}^{2}\frac{4x^2+3x+1}{x(2x+1)}\;{dx} = \displaystyle \underbrace{\int_{1}^{2}\frac{4x^2+3x}{x(2x+1)}\;{  dx}}_{I_{1}}+ \underbrace{\int_{1}^{2}\frac{1}{x(2x+1)}\;{dx}}_{  I_{2}}

    Since we are given that \displaystyle \left[\ln\left(\frac{x}{x(2x+1)}\right)\right]' = \frac{1}{x(2x+1)}, we have \displaystyle I_{2} = \bigg[\ln\left(\frac{x}{x(2x+1)}\right)\bigg]_{1}^{2}.

    For \displaystyle I_{1}, we write \displaystyle I_{1} = \int_{1}^{2}\frac{4x^2+3x}{x(2x+1)}\;{dx} = \int_{1}^{2}\frac{x(4x+3)}{x(2x+1)}\;{dx} = \int_{1}^{2}\frac{4x+3}{2x+1}\;{dx}

    Then let u = 2x+1.
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  6. #6
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    Thanks guys, I've figured out a way to do it already.
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