A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zaxoanl

  • one year ago

need help on integrate dx / (sqrt x)(2+ sqrtx)^4

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    try \(u=\sqrt{x}\)

  2. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Hint: Let \(\LARGE u = 2+\sqrt{x}\) which means \(\LARGE \frac{du}{dx} = \frac{1}{2\sqrt{x}}\) which becomes \(\LARGE 2du = \frac{dx}{\sqrt{x}}\)

  3. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you. i used \[u=2+\sqrt{2}\] and i got \[\frac{ -3 }{ 4 }(2+\sqrt{x)} ^-3 + c\]

  4. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i mean -2/3 (2+.,,,,

  5. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\Large -\frac{2}{3}(2+\sqrt{x})^{-3}+C\] is correct

  6. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is your function: \(\dfrac{dx}{\sqrt{x}}\cdot (2+\sqrt{x})^4\) ?

  7. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you give me a hint on this question as well \[dx/(1+2e^x-e^-x)\]

  8. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its \[dx/(\sqrt{x}(2+\sqrt{x})^4)\]

  9. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    first I would multiply top and bottom by \[\Large e^x\]

  10. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\large \frac{dx}{1+2e^x-e^{-x}}\] \[\large \frac{e^x*dx}{e^x*(1+2e^x-e^{-x})}\] \[\large \frac{e^x*dx}{e^x*1+e^x*2e^x-e^x*e^{-x}}\] \[\large \frac{e^x*dx}{e^x+2e^{2x}-1}\] \[\large \frac{e^x*dx}{e^x+2(e^{x})^2-1}\] \[\large \frac{e^x*dx}{2(e^{x})^2+e^x-1}\]

  11. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Then notice how the denominator is of the form 2z^2 + z - 1 where z = e^x factor 2z^2 + z - 1 to get (z+1)(2z-1) and then use partial fraction decomposition

  12. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i did that and i don't know what number to use to cancel out the e^x

  13. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[e^x = A (e^x + 1) + B (2e^x -1)\]

  14. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\Large e^x = A (e^x + 1) + B (2e^x -1)\] \[\Large e^x = A*e^x + A + 2Be^x - B\] Grouping up the terms, we see that 1*e^x = A*e^x+2Be^x 1*e^x = e^x(A+2B) 1 = A+2B and 0 = A-B

  15. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry don't get how to solve for the x values

  16. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    do you see how to get A and B?

  17. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how would you know what numbers to substitute in to solve for a and b

  18. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait i think i got it

  19. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you saw how I got 1 = A+2B 0 = A-B right? or no?

  20. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i looked at something wrong there you were right so i got \[\ln \left| 2e^x+1 \right|+\ln \left| e^x+1 \right|+c\]

  21. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you're very close, but not quite there

  22. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    You should find that A = B = 1/3

  23. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    also, it's 2e^x - 1 not 2e^x + 1

  24. zaxoanl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yea i wrote it right just that i typed it wrong 2e^x - 1

  25. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{A}{2e^{x}-1}+\frac{B}{e^{x}+1}\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1/3}{2e^{x}-1}+\frac{1/3}{e^{x}+1}\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{1}{2e^{x}-1}+\frac{1}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{1-2e^x+2e^x}{2e^{x}-1}+\frac{1-e^x+e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{-(2e^x-1)+2e^x}{2e^{x}-1}+\frac{e^x+1-e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(-\frac{2e^x-1}{2e^{x}-1}+\frac{2e^x}{2e^{x}-1}+\frac{e^x+1}{e^{x}+1}-\frac{e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(-1+\frac{2e^x}{2e^{x}-1}+1-\frac{e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{2e^x}{2e^{x}-1}-\frac{e^x}{e^{x}+1}\right)\] I'll let you finish up

  26. 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.