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Anti derivative

  1. May 6, 2008 #1
    I just don't get it.

    how does

    the anti derivative of

    ( (20)/(1+x^(2)) )^(2)

    =

    200arctan(x)+(200x)/(x^(2)+1)
     
  2. jcsd
  3. May 6, 2008 #2
    oh p.s. that is how make TI-89 works it out.
     
  4. May 6, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It doesn't. That 200 is wrong.

    The derivative of arctan(x) is 1/(x^2+ 1). The derivative of x/(x^2+ 1), using the quotient rule, is [(1)(x^2+1)- (x)(2x)]/(x^2+ 1)^2= (1- x^2)/(x^2+1)^2

    Their sum is (x^2+ 1)/(x^2+ 1)^2+ (1- x^2)/(x^2+1)^2= 2/(x^2+1)^2. Multiplying by 10, not 200, would give that "20".
     
  5. May 13, 2008 #4
    To get the antiderivative, you could substitute x=tan y and so dx=(sec y)^2 and thus simplifying
    you would end up with (400/sec^2 y) dy
    =>(400cos^2 y) dy
    =>400(1+cos2y)dy
    and integrate this and in the end substitute y=arctan x to get your answer.
     
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