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Homework Help: Evaluate this indefinite integral

  1. Oct 29, 2008 #1
    Evaluate this indefinite integreal

    S = integral

    S 1/(9+x^2)^2

    This has been driving me and my friend nuts.

    We tried partial fractions only to realise that it brings us back to the same thing because its not a polynomial over a polynomial, we tried by parts and it did not help and we tried substitution then trig substitution and uhhhh now were lost
  2. jcsd
  3. Oct 29, 2008 #2


    Staff: Mentor

    Re: antiderivative

    It is a polynomial over a polynomial -- 1 is a polynomial of degree 0. You might not have used the right partial fraction decomposition, which would be
    (Ax + B)/(9 + x^2) + (Cx + D)/(9 + x^2)^2.

    On the other hand, this one is ripe for a trig substitution, I believe, with x/3 = tan(theta). Identify the legs of a right triangle with lengths of x (opposite) and 3 (adjacent). That's the direction I would explore first, if you haven't already done so.
  4. Oct 29, 2008 #3
    Re: antiderivative

    (Ax + B)/(9 + x^2) + (Cx + D)/(9 + x^2)^2

    [ (Ax + B)(9+x^2) + (Cx +D ) ] / ( 9 + x^2 ) ^2 = 1/(9+x^2)^2

    (Ax + B)(9+x^2) + (Cx +D ) = 1

    9Ax + Ax^3 + 9B + B^2 + Cx + D = 1

    Ax^3 + Bx^2 + (9A +C )x + 9B + D = 1

    A=0 , B=0 , C=0 , D=1

    1/(9+x^2) ^ 2 ??

    did i do something wrong
  5. Oct 29, 2008 #4
    Re: antiderivative

    As for trig sub i come down to

    S 1/(U² * sqrt(U-9) )

    which would mean my hypotenus is sqrt(U)
    and i will have one side 3 and the other sqrt( U-9) ? right

    I just never dealt with sqrtU as a side until today
  6. Oct 29, 2008 #5


    Staff: Mentor

    Re: antiderivative

    Nope, your work is correct but not helpful. Partial fraction decomposition didn't break the rational expression into two separate rational expressions.
  7. Oct 29, 2008 #6


    Staff: Mentor

    Re: antiderivative

    Where is your du? When you do a substitution of any kind you need to replace x and dx with u and du.

    With the trig substitution I had in mind, you have

    tan u = x/3, so sec^2 (u) *du = dx/3, or dx = 3*sec^2(u)

    The hypotenuse is not sqrt(u). The relationship is this: sec^2( u ) = (x^2 + 9)/9.

    See if these do you some good.
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