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Math Help - Problem integrating after partial fractions

  1. #1
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    Problem integrating after partial fractions

    (x + 301) / (x^3 + x+ 10)

    Using synthetic division i factored the denominator to (x+2)(x^2 - 2x + 5)

    After the partial fractions technique, A=23 B=-23 and C=93. i resulted with this:

    int(23 / (x+3))dx + int(-23x / (x^2 -2x + 5))dx +int(93 / (x^2 -2x + 5))dx

    i figured out the 1st part using the ln antiderivative. But i cant figure out how to integrate the 2nd 3rd term. Any help is great,
    thanks

    Chris
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  2. #2
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    Hello, Chrs!

    \int\frac{x + 301}{x^3 + x + 10}\,dx


    Using synthetic division, i factored the denominator: . (x+2)(x^2 - 2x + 5) . . . . Good!

    After the partial fractions technique: . A\,=\,23,\;B\,=\,-23,\;C\,=\,93 . . . . Yes!

    i had: . 23\!\!\int\frac{dx}{x+23} \;-\;23\!\!\int\frac{x\,dx}{x^2 -2x + 5} \;+ \;93\!\!\int\frac{dx}{x^2 -2x + 5} . . . . Right!

    i figured out the 1st part using the ln antiderivative.
    But i cant figure out how to integrate the 2nd, 3rd terms.
    Consider them as one fraction: . \frac{-23x + 93}{x^2-2x+5}

    Now we'll perform some Olympic-level gymnastics on the numerator.

    We have: . -23x + 93 \;=\;-23x + 23 + 70

    Factor out -23: . -23\left(x - 1 - \frac{70}{23}\right)

    Muliply by \frac{2}{2}\!:\;\;-\frac{23}{2}\left(2x - 2 - \frac{140}{23}\right)

    And we have: . -\frac{23}{2}(2x-2) + 70


    The integral becomes: . \int\frac{-\frac{23}{2}(2x-2) + 70}{x^2-2x+5}\,dx

    . . =\;-\frac{23}{2}\!\!\underbrace{\int\frac{2x-2}{x^2-2x+5}\,dx}_{\text{This is }\frac{dy}{u}} \;\;+\;\;70\!\!\underbrace{\int\frac{dx}{x^2-2x+5}}_{\text{Complete the square}}


    The last denominator is: . x^2 -2x + 1 + 4 \:=\:(x-1)^2 + 4

    And the last integral is: . 70\!\!\int\frac{dx}{(x-1)^2+4} . . . arctangent form!


    I need a nap!
    .
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  3. #3
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    Hi Soroban, Thank you so much for the help. It really means alot.
    I actually just have one more part that i really just want to confirm, im unsure if im done or not:
    Using the previous findings (the 3 integrals), resolve with the improper integral
    ∫1-->∞ F(x)dx
    That is 1 to infinity, pretty bad looking, i know

    Anyways, for the 3 integrals(ln, ln and arctan) i got
    ∞, ∞, and π/2
    So that would conclude that the integral goes to infinity?
    Is there an answer ti find other than infinity? Sorry for the rediculous fonts.

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  4. #4
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    Hello again, chrsr345!

    The answer is finite . . . but it takes even more gymnastics!


    We had: . 23\int\frac{dx}{x+3} - \frac{23}{2}\int\frac{2x-2}{x^2-2x+5}\,dx + 70\int\frac{dx}{(x-1)^2 + 4}

    Integrate: . 23\ln(x+3) - \frac{23}{2}\ln(x^2-2x+5) + 35\arctan\left(\frac{x-1}{2}\right)


    Factor out \frac{23}{2} from the first two terms:

    . \frac{23}{2}\,\bigg[2\cdot\ln(x+3) - \ln(x^2-2x+5)\bigg] + 35\arctan\left(\frac{x-1}{2}\right)

    . . = \;\frac{23}{2}\cdot\bigg[\ln(x+3)^2 - \ln(x^2-2x+5)\bigg] + 35\arctan\left(\frac{x-1}{2}\right)

    . . = \;\frac{23}{2}\cdot\ln\left(\frac{x^2+6x+9}{x^2-2x+5}\right) + 35\arctan\left(\frac{x-1}{2}\right)


    Now we're supposed to evaluate from 1 to \infty.

    If x=1\!:\;\;\frac{23}{2}\ln\left(\frac{16}{4}\right) + 35\arctan(0) \;=\;\frac{23}{2}\ln(4) + 0 \;=\;\boxed{\frac{23}{2}\ln(4)}


    For "infinity": . \lim_{x\to\infty}\bigg[\frac{23}{2}\ln\left(\frac{x^2+6x+9}{x^2-2x+5}\right) + 35\arctan\left(\frac{x-1}{2}\right)\bigg]

    . . = \;\lim_{x\to\infty}\bigg[\frac{23}{2}\ln\left(\frac{1 + \frac{6}{x} + \frac{9}{x^2}}{1 - \frac{2}{x} + \frac{5}{x^2}}\right) + 35\arctan\left(\frac{x-1}{2}\right)\bigg]

    . . = \;\frac{23}{2}\ln\left(\frac{1+0+0}{1-0+0}\right) + 35\arctan(\infty) \;= \;\frac{23}{2}\ln(1) + 35\left(\frac{\pi}{2}\right) \;=\;\boxed{\frac{35\pi}{2}}


    Therefore, the answer is: . {\color{blue}\frac{35\pi}{2} - \frac{23}{2}\ln(4)}

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