1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: What is the definision of lan transformans

  1. Feb 17, 2007 #1
    i can take every time all the function on the buttom
    put it in a lan
    an devide it in its derivative?

    even if i got a fuction

    its integral solves [(x+2)^-1]/-1

    or i can open this equetion (a+b)^2=x^2+2*x*2+2^2
    then put it all in a lan ln(x^2+2*x*2+2^2) and i will devide it by it derivative



    what is my red sigh
    how do i deside to what way to go???
  2. jcsd
  3. Feb 17, 2007 #2


    User Avatar
    Science Advisor

    I hate to criticize anyone's English (since my (put the language of your choice here!) is awful) but I can't really understand what you are saying here. In particular, what do you mean by " lan"?

    Here is my response to what I think you are asking:
    Yes, it is true that the anti-derivative, or integral, of (x+2)-2 is
    (x+2)-1 divided by -1: that's the power rule: -2+ 1= -1 (and it helps that the derivative of x+ 2 itself is 1).

    If you had (3x+ 2)-2 to integrate, you can do much the same thing: (3x+2)-1 but now divided by the derivative of 3x+ 2 which is 3: (-1/3)(3x+2)-1. That's because if you differentiate (-1/3)(3x+2)-1, you would use both the "power rule" (with n-1= -1-1= -2) and the "chain rule" (the derivative of 3x+2 is 3) to multiply by -1 and 3 and change the power to -2: (-1)(3)(-1/3)(3x+2)-2, the original function.

    However it is important there that the "inner" function, x+ 2 in the first case and 3x+2 in the second, is linear and its derivative is a constant.

    The "chain rule in reverse" is really "integration by substitution": To integrate (3x+2)-2, more formally, let u= 3x+ 2. Then du= 3 dx or dx= (1/3)du and (3x+2)-2= u-2 so [itex]\int (3x+2)^{-2}dx= (1/3)\int u^{-2}du= (-1/3)u^{-1}+ C= (-1/3)(3x+2)^{-1}+C[/itex].

    If that were (x2+ 2)-2 we cannot do that! If we try to let u= x2+ 2 then du= 2x dx and there is NO "2x" in the integral. While we could write (1/3)du= dx before, we cannot write (1/2x)du= dx because we can't have an "x" in the "u" integral.

    In particular, to integrate (x2+ 4x+ 4)-1, which I think is your second example, we CANNOT just say "well since that is a -1 power, the anti-derivative is a logarithm and the we divide by the derivative of x2+ 4x+ 4= 2x" because that derivative is NOT a constant.
    If you tried to make the substitution u= x2+ 4x+ 4, then du= (2x+ 4) dx and we cannot just divide by that!

    Of course, you can always check an integral by differentiating;
    What do you get if you differentiate ln(x2+ 4x+ 4)/(2x+ 4)?

    You would have to use the quotient rule: the derivative of ln(x2+ 4x+ 4) time 2x+4 minus ln(x2+ 4x+ 4) times the derivative of 2x+ 4, all divided by (2x+4)2. That is
    {(1/(x^2+ 4x+ 4)(2x+ 4)(2x+4)- ln(x2+ 4x+ 4)(2)}/(2x+4)2. That is not anything like your original (x2+ 4x+ 4)-1!!

    The rule "integrate f(x)n by integrating un and then dividing by the derivative of f works only if f is linear so its derivative is a constant. Otherwise see if its derivative is already in the integral so you can substitute.
    Last edited by a moderator: Feb 17, 2007
  4. Feb 17, 2007 #3
    so i wich case we know that an integral equals to a ln fuction

    and what are my options in the case of non lenear fraction like x^2+1

    i know that one way is using trigonometric substitution

    but what do i do i a case of e^x for example????
    Last edited: Feb 17, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook