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Need integral help ( integral property maybe )

  1. Oct 4, 2009 #1
    What is the [tex]\int(u)[/tex]-1 where u is a fonction of x , forming a quadratic equation.
    As in:

    [tex]\int(u)[/tex]-1 where u = x2+2ax+a2 for example.

    Is there a basic property for this... or do I have to play around with the fonction u , in order to integrate?


    ----

    I know that [tex]\int(x)[/tex] = ln(x) , but this can't be applied here, right? cuz ln(u)' = u-1(u'). Maybe would I need to find a way to eliminate the u' as a result of the integral.

    OR

    Do I need to change u to = ( x+a)(x+a) and do the integer of that ^-1 , so: [tex]\int((x+a)^-1)((x+a)^-1)[/tex]


    ----

    Does anyone know an integral property for this kind of problem, or know of a way to set me on the right track here?

    Thank you.
     
  2. jcsd
  3. Oct 4, 2009 #2

    mathman

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    Science Advisor

    Since you know the integral of x-2 is -x-1, you should be able to do the integral of (x+a)-2. Hint: let y=x+a and get the relationship between dy and dx.
     
  4. Oct 4, 2009 #3
    Ok, if I get what you're saying: u[tex]\equiv(x+a)[/tex]

    [tex]\int(u)[/tex]-2= -u-1= -(x+a)-1


    I was onto that before, but then... if I do (-u-1)' = u-2u'


    So how come the (u') is not in the equation while doing the integral.
    If I integrate something and then derive it, it should come back to the original term.

    But apparently not cuz: (u-2)u'[tex]\neq(u)[/tex]-2

    **Is that what you meant mathman, or am I still off?
     
  5. Oct 5, 2009 #4

    mathman

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    Since u=x+a then u'=1. Therefore u-2u' = u-2
     
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