Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly Group Title

This is the first dish \[\int \frac{x^2}{1-x^2}dx\] Apart from resolving that into partial fractions, what is a better way to do it?

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    let x=sinu \[\int \frac{x^2}{1-x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2u-cosu)du=tanu-sinu+C\]?

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

    Last line there: tanu -sinu +C \[=\frac{x}{x^2+1}-x+C\] What am I doing here :S

    • 2 years ago
  3. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{sin^2ucosu}{cos^2u}=\frac{(1-cos^2u)cosu}{cos^2u} = sec^2u-cosu\]

    • 2 years ago
  4. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    tan u is x/ sqrt(1-x^2)

    • 2 years ago
  5. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh.. My mistake! \[=\frac{x}{1-x^2} -x+C\]

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

    Alright, let's go to the main dish :S

    • 2 years ago
  7. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    and its not sec^2 , check again

    • 2 years ago
  8. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    sec-cos

    • 2 years ago
  9. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    secu! I'd better do it all over again!!! \[\int \frac{x^2}{1-x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secu-cosu)du\]\[=\ln|secu+tanu|-sinu+C\]

    • 2 years ago
  10. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    not a better way......

    • 2 years ago
  11. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I know, I just try whatever I can.. Is there a better way then?

    • 2 years ago
  12. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't think,partial is best....

    • 2 years ago
  13. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int \frac{x^2}{1-x^2}dx\]\[ = \int (\frac{1}{2(x+1)}-\frac{1}{2(x-1)}-1)dx\]\[=\frac{1}{2} ln|x+1| -\frac{1}{2}ln|x-1|-x+C\]\[=\frac{1}{2}ln|\frac{x+1}{x-1}|-x+C\]

    • 2 years ago
  14. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yup, thats correct.....

    • 2 years ago
    • Attachments:

See more questions >>>

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.