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Help me with this integral

  1. May 19, 2006 #1
    Here is the integral I am suppose to solve for.


    I am using the method of integration by parts.

    Let u = 16/x^2
    dv = 1/(x^2+4) = 1/(x^2 + 2^2)

    Then du = -32/x^3 and v = (1/2)arctan(x/2) ... by use of integral table

    So, then with u*v- int(v*du) ... (16/x^2)*((1/2)arctan(x/2)) - int((1/2)arctan(x/2)*(-32/x^3) dx

    Now I solved the int((1/2)arctan(x/2)*(-32/x^3) dx
    Let u`= (1/2)arctan(x/2)
    dv` = -32/x^3

    Then du` = 1/(x^2 + 2^2) and v`=16/x^2

    So I go on to solve this integral (1/2)arctan(x/2)*(16/x^2) - int(16/((x^2)(x^2+4))

    The problem is when I add like terms to solve the whole integral, the term (1/2)arctan(x/2)*(16/x^2) cancels out and here is where I get stuck. I know that I am doing something wront with the above math but I just can't seem to find it.

    Their is a lower limit of 2 and an upper limit of 4 but that is irrelevant for now.
  2. jcsd
  3. May 19, 2006 #2


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    What you did was integrate by parts twice: the second application of it undid the first, hence you are left with the integral you started with. Use instead partial fraction decomposition, namely use the fact that

  4. May 19, 2006 #3


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    Just a little typo here, benorin, the LaTeX line should read:
    [tex]\frac{16}{x ^ 2 (x ^ 2 + 4)} = \frac{4}{x ^ 2} - \frac{4}{x ^ 2 + 4}[/tex] :wink:
  5. May 19, 2006 #4
    You are using integration of parts twice here, once in one direction- using 1/x as u to get du of -1/x^3, and then using -1/x^3 as dv, finding 1/x^2 as v. Essentially, you are going in circles.

    You may find partial fraction decomposition an easier method prior to trying to integrate.
  6. May 21, 2006 #5


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    Thank you VietDao29.
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