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Integration by parts?

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the indefinite integral.

    ∫(x + 3)/(x^2+6x) dx

    2. Relevant equations
    This is an online hw prob. that covers sections integration by parts and substitution in indefinite integrals. it looks to me that it fits into the formula ∫udv=uv-∫vdu if you change the original equation to (x+3)*(x^2+6x)^-1

    3. The attempt at a solution
    Doing this would give you u=x+3, dv=(x^2+6x)^-1, du=1, v=(-log(x+6)-log(x))/6

    the final answer becomes (x+3)((-log(x+6)-log(x))/6)-((-(x+6)log(x+6)+xlog(x)-2x-6)/6)

    this however is not correct answer. Am I using the wrong method?
  2. jcsd
  3. Jun 9, 2010 #2
    You are missing an easier method. Take the derivative of the denominator and compare it to the expression in the numerator. Do you see an easy relation between the TWO? Now ask, what substitution makes sense?
  4. Jun 9, 2010 #3


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    I think you'd be better served trying substitution.
  5. Jun 9, 2010 #4
    Are you suggesting sub u=x^2+6x, du=2x+6 dx and so 1/2du=x+3 dx and the final answer should be 1/2 log (x^2+6x)? This is not the answer he wants because i believe he wants us to use a different method. Kind of confusing.
  6. Jun 9, 2010 #5
    It is confusing. Did he say what method he wants you to use? You mentioned the problem came from an online HW problem directed at int. by parts and substitution. It seems substitution is the preferred and direct method. But, there is usually more than one way to skin a cat.
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