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whatisthequestion
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Evaluate the integral. The integral will be entered below. The title of this section is Intergration of Rational Fractions by Partial Fractions
 one year ago
 one year ago
whatisthequestion Group Title
Evaluate the integral. The integral will be entered below. The title of this section is Intergration of Rational Fractions by Partial Fractions
 one year ago
 one year ago

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whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ x^2x+6 }{ x^3+3x }dx\]
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix does this one only simplify down to \[\int\limits_{}^{}\frac{ x^2x+6 }{ x(x^2+3) }dx\] 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?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{x^2x+6}{x(x^2+3}\qquad = \qquad \frac{A}{x}+\color{royalblue}{\frac{Bx+C}{x^2+3}}\] 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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
You `could` factor the \(\large x^2+3\) into complex conjugates. But I think that would get kinda messy :) Definitely not worth the effort.
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
okay so then its \[x^2x+6=A(x^2+3)+Bx(x)+C(x)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Looks good c:
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The difference of squares can be broken into conjugates right? \(\large x^2a^2 \qquad = \qquad (xa)(x+a)\) 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). \(\large x^2+a^2 \qquad = \qquad (x+ai)(xai)\)
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Ah I see :)
 one year ago
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