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An integral with a fraction in it-need help finding the antiderivative

  1. May 6, 2008 #1
    An integral with a fraction in it--need help finding the antiderivative

    1. The problem statement, all variables and given/known data
    I have the integral [tex]\int(\frac{20}{1+x^2} - 2)dx[/tex] where x = ± 3. To solve, I need to find the antiderivative of [tex]\frac{20}{1+x^2} - 2[/tex]

    Now, the worksheet actually gives the answer to the problem: The antiderivative is 37.961 or 37.962. However, I'm having trouble actually reaching that answer on my own; I have trouble with the antiderivatives of fractions.

    2. Relevant equations

    Lessee... I know that the antiderivative of 1/x is ln x. When it gets any more complex than that I get confused.

    3. The attempt at a solution

    First, I took [tex]\int(\frac{20}{1+x^2} - 2)dx[/tex] and turned it into [tex]\int(20(1+x^2)^-^1 - 2)dx[/tex]

    Then, I tried to take the antiderivative. I thought it would be: [tex]20ln(1+x^2)-2x.[/tex]

    Then I plugged in the x values, 3 and -3. With positive 3 plugged in, it came out to 40.052, with a negative 3 it came out to 52.052. I don't even have to finish it off with the subtraction to know that my final answer isn't right and that I missed a step (or two) somewhere. Only question is: What step did I miss? I have a feeling I was supposed to do something with the 20... And I think that there might have been more to taking the antiderivative of (1+x^2)^-1 then just turning it into a natural logarithm. Would someone please help me figure out what I forgot?
  2. jcsd
  3. May 6, 2008 #2


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    Try a trig sub or look up derivatives of inverse trig functions.
  4. May 6, 2008 #3


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    Homework Helper

    If you're using a trigo substitution, you may want to use this:
    [tex]tan^2\theta + 1 = sec^2\theta[/tex]

    That should help you.
  5. May 6, 2008 #4
    Don't be a victim of universal logarithmic differentiation!

    [tex]\int\frac{20}{1+x^2}dx[/tex] doesn't equal [tex]20*ln|1+x^2|+c[/tex]
    Last edited: May 6, 2008
  6. May 6, 2008 #5
    If you haven't touched trig int before:

    let x = tanu
    dx = tanu.secu

    substitute these values for x and dx
  7. May 6, 2008 #6
    Oh hush! We all know it's 20arctanx - 2x! Cut the kid some slack!

    The reason the natural logarithm didn't work was because you need the derivative of the denominator in the numerator. The derivative of 1+x^2 is 2x, and as you can see, it ain't there.
  8. May 6, 2008 #7
    I didn't know it three-four years ago :shy:...:cry:
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