1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

    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

    Mark44

    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

    Mark44

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Evaluate this indefinite integral
Loading...