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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


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    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: 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
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