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Fun problem: ? x / (x^2 + 6x + 10) dx

  1. Mar 17, 2004 #1
    Fun problem: ? x / (x^2 + 6x + 10) dx

    Integration by parts proves 1=1! My mathematical fame is at hand! So how would you do this one?
  2. jcsd
  3. Mar 18, 2004 #2
    [tex]\int\frac{x}{x^2 + 6x + 10}{\rm d}x = \frac12\left(\ln[10 + x(6 + x)]-6 \arctan [3 + x] \right)[/tex]

    So, what does this have to do with 1=1 (which is selfevidently true anyway)?
    Last edited: Mar 18, 2004
  4. Mar 18, 2004 #3


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    Integration by parts proves 1=1? In other words, you used integration by parts twice, the second time reversing your choice for u and dv so the two cancelled!

    "Partial fractions" is what you need here. The denominator, x^2 + 6x + 10, is "irreducible" over the real numbers. It is the same as
    x^2+ 6x+ 9+ 1= (x+3)^2+ 1. I would recommend the substitution
    u= x+ 3 so that du= dx, x= u- 3 and the problem becomes integrating
    (u-3)/(u^2+1)= u/(u^2+1)- 3/(u^2+1).

    The first of those can be done by the further substitution v= u^2+1 and the second is a simple arctangent.
  5. Mar 18, 2004 #4
    OOops, I forgot that x could be expressed in terms of u.
    Last edited: Mar 19, 2004
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