Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

whatisthequestion

  • 2 years ago

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

  • This Question is Closed
  1. whatisthequestion
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. whatisthequestion
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \frac{x^2-x+6}{x(x^2+3}\qquad = \qquad \frac{A}{x}+\color{royalblue}{\frac{Bx+C}{x^2+3}}\] 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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  5. whatisthequestion
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Looks good c:

  7. whatisthequestion
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  8. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  10. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ah I see :)

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.