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Is this integral formula correct?

  1. Jan 9, 2016 #1
    Quick question: is the following a correct integral formula?
    $$\int \! \frac{1}{(-x + a)^2} dx = -\left(\frac{1}{-x + a}\right)$$
    I couldn't find it on the integral tables, so I just extrapolated from a similar integral formula I've found there. That formula being:
    $$\int \! \frac{1}{(x + a)^2} dx = -\left(\frac{1}{x + a}\right)$$
    Now ladies and gentlemen...if my extrapolation was incorrect, would you be so generous as to provide me with the correct formula? Or share with me the proper methodology as to how to generate it, if my quick and dirty guesswork is inappropriate?
    o_O
     
  2. jcsd
  3. Jan 9, 2016 #2

    PAllen

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    You have a sign error (you are correct up to sign). You can see what the correction should be, why it is needed, and verify any such guess by taking the derivative. I will leave that to you.
     
  4. Jan 9, 2016 #3

    fresh_42

    Staff: Mentor

    They are correct.
    EDIT: PAllen is right. I overlooked your minus. Hint: Use the chain rule of differentiation.
    EDIT2: I hate to say this, but you have a problem concerning the constants!
     
  5. Jan 9, 2016 #4
    fresh_42, I looked at the link you had given, but can't understand why it proves your assertion that my extrapolated formula is correct. Can you explain?:confused:
     
  6. Jan 9, 2016 #5

    fresh_42

    Staff: Mentor

    Your first formula is not quite correct. Differentiate it by using the chain rule and you will see.
    The page I gave you lists your formula as third in its chapter "Alternative Forms" ##(n=-2)##.
    This page here guides you to an awful lot of formulas for integration. However, to get used to and to develop a feeling for them it makes sense to control them by differentiation before you use them. Plus Wikipedia might have typos and like you it doesn't always contain the constant. I know the constant seems to be annoying but you better get used to it. You will avoid a lot of mistakes in more complicated situations. E.g. it played a crucial role in yesterdays solution, at least the long one. Once you're firm in the subject you may be sloppy with it but only then.
     
  7. Jan 9, 2016 #6

    PAllen

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    Yes, it is correct to include constants in an indefinite integral expression, however if you are only going to use the formula to compute a definite integral, it does not matter. I chose to ignore that issue, as the sign error (and the apparent failure to check using differentiation) were the substantive issues.
    You shouldn't be relying on tables of integrals if you can't check via differentiation. Getting the integral can be hard, but checking should be easy. In your case, checking is very easy (finding you are correct up to sign .. and constant, for generality).
     
  8. Jan 9, 2016 #7
    After reviewing what you both were speaking of, I believe I understand, as I shall show below...

    I begin by taking the formula I had devised, and restating it thusly, as well as setting it to f(x):
    $$-\left(\frac{1}{-x+a} + C\right) = -(-x + a)^{-1} - C = f(x)$$
    Next, to find the derivative of f(x), I apply the chain rule by differentiating first the outside (which causes the constant to disappear):
    $$(-1) *(- (-x + a)^{-2}) - 0 = \frac{1}{(-x+a)^2}$$
    And then I differentiate the inside (causing "a" to disappear):
    $$(1)(-x)^{0} + 0 = -1$$
    Then I multiply these two together, because the chain rule demands that the outside derivative is multiplied by the inside derivative:
    $$\left(\frac{1}{(-x+a)^2}\right)*(-1) = -\left(\frac{1}{(-x+a)^2}\right) \neq \left(\frac{1}{(-x+a)^2}\right)$$
    Therefore, the integral formula is actually:
    $$\int \frac{1}{(-x+a)^2} dx = \frac{1}{-x+a}$$

    Thank you, comrades. This has been altogether a most rewardable lesson throughout this thread.
    :cool:
     
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