Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

eistein+newton+gallileo=_me

  • 4 years ago

please help me solve these using u substitution ( please show the workings) 1. e^x(root(1+e^x)) 2.xtan(x^2) 3 sin(rootx)/rootx

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

    Are you looking to take a derivative?

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

    My guess would be the integral

  3. eistein+newton+gallileo=_me
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please integrate them through u subs

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

    OK, which one are u doing tbates?

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

    the second one I have

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

    1. u = 1+e^x du = e^x dx so \[\int\limits{\sqrt{u}du}\]

  7. lalaly
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2. u= x^2 du= 2xdx \[\frac{1}{2} \int\limits{tanu du}\]

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

    On the second one set u=x^2 du = 2x dx du/2=2xdx Sub in: tan(u)/2 du The integral of tan(u)/2 -(1/2)ln|cos u| + C Plug back in for x -(1/2)ln|cos x^2| + C

  9. eistein+newton+gallileo=_me
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanx guys and number 3?

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

    u= sqrtx du = 1/(2sqrtx) \[2 \int\limits{sinu du}\]

  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