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Integration by parts

  1. Dec 27, 2008 #1
    1. The problem statement, all variables and given/known data

    integrate ln(x+1) dx (integrate by part)

    2. Relevant equations
    integrate e^-2x sin3x dx


    3. The attempt at a solution
    1st,i make u=x+1 ,so du/dx =1 du=dx..while i use /dv =/ln(x+1) dx and after integrating my v=1/(x+1) so subtitute into uv -/vdu but my ans turn out to be wrong so i need some guidance over here.i guess my integration for dv is wrong but i can't really see it..so could any1 please help me..thanks..:blushing:
     
  2. jcsd
  3. Dec 27, 2008 #2
    How about letting u = ln(x+1) and dv = 1 dx?

    I'm not sure what you meant by letting u = x + 1 and dv = ln(x+1). You have integral of ln(x+1) dx, you are subst. u and dv in there, once you let u = x + 1 it makes no sense to let dv = ln(x+1)
     
  4. Dec 27, 2008 #3
    uv -/vdu

    /dv =/ln(x+1) dx

    if / is integral sign.. then that ln(x+1) is wrong.

    It's like

    d/dx ( x*ln(x+1)) = ?

    and uv = x*ln(x+2)

    You just rearrange that "d/dx ( x*ln(x+1)) = ?" equation .. to find int [ln(x+1)] .. I never learned using u and v

    Edit: I think NoMoreExams is right... I never used u-v formula
     
  5. Dec 27, 2008 #4
    i did like what you mention but i couldn't find the ans..the ans is (x+1)(ln(x+1)) -(x+1) + c ,if i differetiate du/dx =1/(x+1) and sub into the formula i can't get..
     
  6. Dec 27, 2008 #5
    ya that's integral sign,why do you say it's wrong?can you please tell me..?thanks..i did like what nomoreexams said..i still can't get my ans correct..i sow you my steps please rectify me.. like what he says; u=ln(x+1) after diff du/dx = 1/x+1 and u integrate dv=1 dx..it will be v= x so finally sub into the formula integration by parts will be uv-/vdu =(ln(x+1))(x)-/(x)(1/x+1 dx) but the ans turn out to be wrong..
     
  7. Dec 27, 2008 #6
    uv-/vdu =(ln(x+1))(x)-/(x)(1/x+1 dx) is NOT equal to (x+1)(ln(x+1)) -(x+1) + c

    integral of x/(x+1) is wrong. Are you using partial fractions here?

    Edit Sorry,:
    >> int (log(x+1),x)

    ans =

    log(x+1)*(x+1)-x-1

    I guess I was making some mistake in the partial fractions.. but you answer looks correct (that's matlab). And, also you can include 1 in c .. it really doesn't make any difference.
     
    Last edited: Dec 27, 2008
  8. Dec 27, 2008 #7
    ya,the answer is not match then which part of mine turn out to be wrong?then integrate of 1/(x+1) wrong?why?so what should i do?no,que required to integrate by parts,which means
    /u dv = uv - /v du
     
  9. Dec 27, 2008 #8
    You guys should look into learning how to typeset in TeX. In any case I'll start you off

    [tex] \int{ln(x+1) dx} [/tex] is what we have.

    If we let [tex] u = ln(x+1) \Rightarrow du = \frac{dx}{x+1} [/tex] and [tex] dv = dx \Rightarrow v = x [/tex]

    Subst. this into our formula we get

    [tex] \int{ln(x+1) dx = x ln(x+1) - \int \frac{x}{x+1} dx [/tex]

    Can you take it from here?
     
  10. Dec 28, 2008 #9
    ok,thanks a lot.,i get your idea..but i can't really take it from there but then i can get it by putting your idea into my way..anyway thank for your help
     
  11. Dec 28, 2008 #10
    You can't integrate x/(x+1) dx?
     
  12. Dec 28, 2008 #11
    Yes, I think he had "(x+1)(ln(x+1)) -(x+1) + c " which is right

    diff {(x+1)(ln(x+1)) -(x+1) + c }
    ln(x+1) + 1 -1 = ln(x+1)

    so which is correct. I don't see any problem here other than 1 which can be included in the constant ...
     
  13. Dec 28, 2008 #12
    I didn't disagree. I was commenting on him saying "but i can't really take it from there" which meant he couldn't integrate x/(x+1) dx.
     
  14. Dec 28, 2008 #13
    i could integrate it thanks a lot,yesterday was brain dead..im so sorry,finally i get your points clear d..thanks man,integrate /x/(x+1) will be ln(x+1) i completely understand..thanks to both of you..but i put it in this way,can you please check out could this sol works? 1st,i use a=x+1 ,then integrate lna da i make u=ln a while dv=da so after du/dx=1/a while v=a..from here i depart to sub into the formula uv-/vdu so it will be ...(lna)(a) - /(a) X (1/a da)
    then it finally turn out to be ln(x+1)(x+1) - (x+1) + c... does this way work?please do rectify me if there is any mistakes i have made...
     
  15. Dec 28, 2008 #14
    No... integral of x/(x+1) dx is not ln(x+1), that would be integral 1/(x+1) dx.

    If you wanted to work with integral of ln(a) da instead of ln(x+1) dx, you can convert it as you said, I don't see why you would but you can.
     
  16. Dec 29, 2008 #15
    oh..ok..im still very weak in integration by parts..do you think of anything which could improve me on that topic?
     
  17. Dec 29, 2008 #16
    Practice :) Once you do hundreds of these problems, you'll be able to catch on faster and at least do a few steps in your head. There's also something called tabular integration which you can look up, it simplifies a certain type of integration by parts problems.
     
  18. Dec 29, 2008 #17
    oh,ya sure i tried a lot of these kind of problems..and catching up well,do you have any link which leads to more ques with solutions?by the way thanks for your help,if not i won't be able to get these things solved..haha..
     
  19. Dec 29, 2008 #18
    which book you are using?
    Most calculus books have like +200 integration questions ... You can always go to the library and find Stewart etc.
    http://www.stewartcalculus.com/

    I remember looking for some practice sites but wasn't really successful.

    And, you don't need to know the answers for all the questions you solve.
     
  20. Dec 29, 2008 #19
    ok,thank you...where are you from?i'm using calculus university as my reference..
     
  21. Dec 29, 2008 #20
    I think Stewart is popular in US/Canada. I was using Glyn James book (a British), but I just went to my library and borrowed some 6 calculus books and then worked on like 2 books (one was Stewart).
     
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