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Using the u of substitution

  1. Apr 12, 2013 #1
    ∫1/(10p-p^2)dp

    i tried using the u of substitution but for some reason I am unable to isolate dp and get an equation in terms of du which i could then plug into the integral and take the antiderivative.
     
  2. jcsd
  3. Apr 12, 2013 #2

    CompuChip

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    Since there is a p2 in the denominator, the first thing that comes to mind is
    [tex]\int \frac{1}{1 + p^2} \, dp = \arctan(p)[/tex]

    So I would try completing the square, i.e. write it as
    [tex]k \cdot \int \frac{1}{1 + u^2} \, du[/tex]
    where u = (p + a) / b.
     
  4. Apr 12, 2013 #3
    not quite sure where you got u=(p+a)/b
     
  5. Apr 12, 2013 #4
    do you mean i should factor out a 10 from denominator which would give me:

    (1/10)∫1/(p-(p^2)/10)dp
     
  6. Apr 12, 2013 #5

    CompuChip

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    No, start by writing 10p - p2 as (p - a)2 + b.
     
  7. Apr 12, 2013 #6

    CompuChip

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    By the way, I was assuming that you know the arctan integral. If not, you could try splitting fractions, i.e. writing
    [tex]\frac{10p - p^2} = \frac{A}{10p} - \frac{B}{p^2}[/tex]
     
  8. Apr 12, 2013 #7
    I would like get this into a Ln function instead of using arctan
     
  9. Apr 12, 2013 #8
    are you using a partial fraction decomp in this case. if so that would be my preferred choice at getting to a solution.
     
  10. Apr 12, 2013 #9
    when i find the perfect square how do i find what the perferct square would be. I dont even know where to start. i punched the problem into wolfram and it completed the square of denom by turning

    1/(10p-p^2) which i guess is equal to 1/(25-(x-5)^2), how do i do this?
     
  11. Apr 12, 2013 #10

    CompuChip

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    OK in that case first try with the partial fraction decomposition, once you have arrived at the solution I will show you the method I was thinking of initially for completeness.
     
  12. Apr 12, 2013 #11

    Mark44

    Staff: Mentor

    I'm sure you mean
    $$ \frac{1}{p(10 - p)} = \frac{A}{p} + \frac{B}{10 - p}$$
     
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