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Integral - just need checking

  1. Jan 27, 2006 #1
    Am I right in saying that the integral of 1/(x-6)^2 dx

    = (x-6)^-1 / -1

    = - 1 / (x-6)
  2. jcsd
  3. Jan 27, 2006 #2


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    Add the Integration constant, and you are correct.
  4. Jan 27, 2006 #3
    I will thanks very much :-)... oh could someone explain exactly what this integral constant is? And why is it needed?
  5. Jan 27, 2006 #4


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    When we write [itex]\int x dx[/itex] we're looking for the function(s) which give x again if we take the derivative. Now, you know this integral is x²/2 since (x²/2)' = x but when you add a constant, it's still valid: (x²/2 + c)' = x. This is because the derivative of a constant is zero.
  6. Jan 27, 2006 #5
    thanks TD for your explanation
  7. Jan 27, 2006 #6


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    You're welcome :smile:
  8. Jan 27, 2006 #7


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    If you have a function

    f(x) = x^2 there are other functions of the same shape that difer only in the position respect "y"...

    if f(x)=x; is like if you have f(x)=x+0=x+C, the +C let yo translate up or down the graph of the function.

    Is necesary when integrating a function, because you dont know wich of this infinite posible values to C is the one, thus, instead of write a infinite set of functions that only difer in the a real number "you select" one specific function of those many posibles that represent such infinite funtions and call it F(x)+C, thus ;), you have one function that can represent any function with the same shape and that only difer in the y position instead of have a infinite number of functions to analise.
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