Question

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

Answer 1

\[\int\limits_{}^{}\frac{ x^2-x+6 }{ x^3+3x }dx\]

Answer 2

@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?

Answer 3

\[\large \frac{x^2-x+6}{x(x^2+3}\qquad = \qquad \frac{A}{x}+\color{royalblue}{\frac{Bx+C}{x^2+3}}\]

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.

Answer 4

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

Answer 5

okay so then its
\[x^2-x+6=A(x^2+3)+Bx(x)+C(x)\]

Answer 6

Looks good c:

Answer 7

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.

Answer 8

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).
\(\large x^2+a^2 \qquad = \qquad (x+ai)(x-ai)\)

Answer 9

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.

Answer 10

Ah I see :)