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whatisthequestion

  • 3 years ago

Evaluate the integral. The integral will be entered below. The title of this section is Intergration of Rational Fractions by Partial Fractions

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  1. whatisthequestion
    • 3 years ago
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    \[\int\limits_{}^{}\frac{ x^2-x+6 }{ x^3+3x }dx\]

  2. whatisthequestion
    • 3 years ago
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    @zepdrix does this one only simplify down to \[\int\limits_{}^{}\frac{ x^2-x+6 }{ x(x^2+3) }dx\] or am I making a silly mistake somewhere again. If that is as far as it factors down does the (x^2+3) change anything?

  3. zepdrix
    • 3 years ago
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    \[\large \frac{x^2-x+6}{x(x^2+3}\qquad = \qquad \frac{A}{x}+\color{royalblue}{\frac{Bx+C}{x^2+3}}\] With partial fractions, you're top part will be one degree less than the bottom. See how the first fraction is a constant over x^1? Our second fraction will be `linear` on top, since the bottom is a quadratic.

  4. zepdrix
    • 3 years ago
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    You `could` factor the \(\large x^2+3\) into complex conjugates. But I think that would get kinda messy :) Definitely not worth the effort.

  5. whatisthequestion
    • 3 years ago
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    okay so then its \[x^2-x+6=A(x^2+3)+Bx(x)+C(x)\]

  6. zepdrix
    • 3 years ago
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    Looks good c:

  7. whatisthequestion
    • 3 years ago
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    could you quickly remind me what a complex conjugate is though? It sounds familiar but I don't know for sure if I have used them before.

  8. zepdrix
    • 3 years ago
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    The difference of squares can be broken into conjugates right? \(\large x^2-a^2 \qquad = \qquad (x-a)(x+a)\) The `sum` of squares produces complex conjgates. It might become relevant if you take differential equations, or complex analysis (I dunno, I haven't gotten that far yet myself). \(\large x^2+a^2 \qquad = \qquad (x+ai)(x-ai)\)

  9. whatisthequestion
    • 3 years ago
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    Okay thats the high level stuff my professor randomly talked about the other day then said we didn't need to know about. I knew it sounded familiar.

  10. zepdrix
    • 3 years ago
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    Ah I see :)

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