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Find the antiderivative

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    find the antiderivative of 1/(x^2)


    3. The attempt at a solution

    I'm pretty sure you just find the antiderivatives of the numerator and the denominator.

    the antiderivative of 1 is x.
    the antiderivative of x^2 is (x^3)/3
    mutliply the numerator by the inverse of the deominator and you get 3/(x^2)

    The book says the answer is -1/x
     
  2. jcsd
  3. Feb 2, 2012 #2

    Ray Vickson

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    Have you ever seen the result ∫ x^n dx = x^(n+1)/(n+1) + C? It does not hold for n = -1 because that would involve division by zero on the right-hand-side, but otherwise n is unrestricted.

    Please, please, please *get rid of the idea forever* that you can do the integral by integrating the numerator and denominator separately: that does not work! For example, integrate x, to get x^2/2. Now write x as x^2 / x and try your method of integrating the numerator and denominator separately---you will get the wrong answer.

    RGV
     
    Last edited: Feb 2, 2012
  4. Feb 2, 2012 #3
    ... (posted an idea that was false, subsequently deleted it)
     
  5. Feb 2, 2012 #4
    Then how do I find the antiderivative of a fraction?

    Here's another problem

    1. 2/3 sec^2 x/3

    2. (2/3 tan x/3)/(x/3)

    3. (2/3 tan x/3) * (3/x)

    4. 2/3 * 3/x (tan x/3)

    5. 2/x tan x/3

    the book says the answer is

    2 tan (x/3)
     
  6. Feb 2, 2012 #5
    Ok, I saw from another website, that to find the antiderivative of a fraction you have to convert the fraction into a number with an exponent.

    so -1/x = -x^-1

    apply (x^n+1)/(n+1)

    that comes to

    -(x^0)/0
     
  7. Feb 2, 2012 #6

    SammyS

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    Assuming that you mean to find the anti-derivative of (2/3) sec2 (x/3) :

    Rather than trying to follow those steps, which I can't make sense of ... Let's see if the book's answer is correct.

    If 2 tan(x/3) is an anti-derivative of (2/3) sec2 (x/3), then the derivative of 2 tan(x/3), with respect to x, should be (2/3) sec2 (x/3) .

    [itex]\displaystyle \frac{d}{dx}\tan(x)=\sec^2(x)[/itex]

    So that: [itex]\displaystyle \frac{d}{dx}\left(2\,\tan\left(\frac{x}{3}\right) \right)=2\,\sec^2\left(\frac{x}{3}\right)\cdot \frac{1}{3}=\frac{2}{3}\, \sec^2\left(\frac{x}{3}\right)\,.[/itex]

    The book's answer is correct.
     
  8. Feb 2, 2012 #7
    Knowing the answer than finding the derivative can be done, but I still don't see how to find the answer starting from the antiderivative.

    I got the answer to 1/x^-2 now.
     
  9. Feb 2, 2012 #8

    SammyS

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    The anti derivative of xn=xn+1/(n=1), except if n = -1.

    Why that exception?
    It's because x0=1, and the derivative of 1 is 0, not x-1.​

    The derivative of x-1 = ln(|x|)
     
  10. Feb 2, 2012 #9

    SammyS

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    The anti-derivative of 1/x-2 is easy, because 1/x-2 = x2.

    Earlier, you wanted the anti-derivative of 1/x2, which you found to be -1/x . That's easy to get if you write 1/x2 as x-2 .

    Details ... details.
     
  11. Feb 2, 2012 #10

    Ray Vickson

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    Did you fail to read my first post, which specifically told you this result fails for n = -1?

    RGV
     
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