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Help me with this integral

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

    int(16/((x^2)(x^2+4))

    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

    benorin

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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

    [tex]\frac{16}{x^2(x^2+4)}=\frac{4}{x^2}-\frac{4}{x^2(x^2+4)}[/tex]
     
  4. May 19, 2006 #3

    VietDao29

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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

    benorin

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    Thank you VietDao29.
     
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