1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral of t/1+t^2

  1. May 19, 2008 #1
    hello there,im having trouble with this question..
    find int t/1+t^2 dt

    my answer is 1/2 ln t (1+t^2)+c
    but the answer that i have copy is 1/2 ln (1+t^2)+c
    maybe i have copy it wrongly,can someone tell me which one is right?
    thx.. :smile:
     
  2. jcsd
  3. May 19, 2008 #2

    exk

    User Avatar

    How did you get your answer?
     
  4. May 19, 2008 #3
    im using the integration using substitution..
    im confuse right now.. can anyone give me the full calculation or what method i should use.
     
  5. May 19, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    [tex]\frac{t}{1+t^2} dt[/tex]

    Let [itex] u=1+t^2[/itex] what is [itex]\frac{du}{dt}[/itex]? then what is dt in terms of du?
     
  6. May 19, 2008 #5
    here it is,pls correct me if im wrong.
    int t/1+t^2 dt
    u = 1+t^2
    du/dt = 2t
    dt = 1/2t du
    = int (t/u)(1/2t du)
    = int t/2tu du
    = 1/2 int t/tu du
    = 1/2 int t(tu)^-1 du
    = 1/2 int t^0 u^-1 du
    = 1/2 [t^1 ln u]+c
    = 1/2 ln t (1+t^2)+c (answer)
     
    Last edited: May 19, 2008
  7. May 20, 2008 #6
    could anyone help me here? pls..
     
  8. May 20, 2008 #7

    Defennder

    User Avatar
    Homework Helper

    Cancel out t in this step.
     
  9. May 20, 2008 #8
    u mean remove the t? so it will become 1/2u?
    here,
    = int 1/2u du
    = int u^-1/2 du
    = 1/2 int u^-1 du
    =1/2 [ln u]+c
    = 1/2 ln (1+t^2)+c (answer)
    am i correct? btw if we remove t,it will become 1/2u right since it was 1t/2tu?
     
  10. May 20, 2008 #9

    Defennder

    User Avatar
    Homework Helper

    Yes, that is correct.
     
  11. May 20, 2008 #10
    thank you! :)
    i got one more question,
    differentiate y= 2m e^mt/cos 2m
    is it dy/dt or dy/dm?
    2m e^mt,
    e^mt if differentiate = e^mt or me^mt?
     
  12. May 20, 2008 #11

    Defennder

    User Avatar
    Homework Helper

    I can't read what you are writing here, is it supposed to be [tex]y = \frac{2me^{mt}}{cos(2m)}[/tex]?

    Is m a constant or function of t?

    As for your 2nd question, it is true that [tex]\frac{d}{dt} \ e^{mt} = me^{mt}[/tex].
     
  13. May 20, 2008 #12
    it is correct except the cos have no bracket, cos2m
    i have no idea,im having trouble to decide whether it is dy/dm or dy/dt
     
  14. May 20, 2008 #13

    Defennder

    User Avatar
    Homework Helper

    Is m a function of t? Secondly what are you differentiating with respect to?
     
  15. May 20, 2008 #14
    here is my answer,do tell whether it is right or wrong.
    2e^mt(m cos 2m+cos 2m+2m sin 2m)/(cos 2m)^2
     
  16. May 20, 2008 #15
    just assume it as dy/dt
    sorry im not good in english.
    im using quotient rule to solve that.
     
  17. May 21, 2008 #16

    Defennder

    User Avatar
    Homework Helper

    You haven't clarified this question: Is 'm' a constant? Or is it a function of t?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integral of t/1+t^2
  1. Integrate t.sin(t^2-1) (Replies: 15)

Loading...