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Troublesome Integral

  1. Mar 16, 2014 #1
    Hello,

    I have tried computing the following integral three times now, and I can not figure out what I am doing wrong.

    The integral is [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh[/itex], which wolfram calculates as [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 (-h - \frac{100}{h-10} + c )[/itex].

    Here is one sample of my work:

    Using integrating by parts, let [itex]u = h^2 - 20h[/itex], which becomes [itex]\frac{du}{dh} = 2h - 20[/itex]. Multiplying both sides by the differential [itex]u[/itex], we get [itex]\frac{du}{dh} dh = (2h-20)dh[/itex]. The definition of the differential of [itex]u[/itex] is [itex]du = \frac{du}{dh} dh[/itex]. Making this substitution, [itex]du = (2h - 20) dh[/itex]. Let [itex]dv = \frac{dh}{(h-10)^2}[/itex]. Integrating both sides, [itex]v = \frac{1}{h-10}[/itex].

    [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[ (h^2-20h) \left(\frac{1}{h-10} \right) - \int \frac{2h-20}{h-10} dh \right] \implies[/itex]

    [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int \frac{h-10}{h-10} dh \right] \implies [/itex]

    [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int dh \right] \implies[/itex]

    [itex]100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2h + c \right] [/itex]

    which is not the same...

    Can anyone see what I might have done wrong?
     
    Last edited: Mar 17, 2014
  2. jcsd
  3. Mar 16, 2014 #2

    Dick

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    They differ by a constant. Divide ##\frac{h^2-20h}{h-10}## out using polynomial division. Using partial fractions to begin with would have led to an easier integration problem. BTW both solutions look to have a sign error.
     
    Last edited: Mar 16, 2014
  4. Mar 16, 2014 #3
    Try using the substitution $$u = h-10$$
    After some simple algebra, you should end up with the integral $$100∫\frac{u^2-100}{u^2} du$$ which should be fairly easy to evaluate, then you can substitute your h-10 back into your solution.
     
    Last edited: Mar 16, 2014
  5. Mar 16, 2014 #4

    lurflurf

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    try using
    $$100 \int \! \frac{h^2-20h}{(h-10)^2} \, \mathrm{d}h=100 \int \! \frac{(h-10)^2-100}{(h-10)^2} \, \mathrm{d}h$$
     
  6. Mar 16, 2014 #5

    Dick

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    Your particular mistake was integrating ##dv = \frac{dh}{(h-10)^2}## to get ##v = \frac{1}{h-10}##. That's pretty wrong. I don't know what Wolfram Alpha's problem is if you quoted it correctly.
     
    Last edited: Mar 16, 2014
  7. Mar 17, 2014 #6
    I wouldn't consider that pretty wrong--perhaps just off the mark; after all, I am just missing a mere minus sign.
     
  8. Mar 17, 2014 #7

    Dick

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    Why is the Wolfram Alpha answer you quoted also off by the same sign?
     
  9. Mar 17, 2014 #8

    vanhees71

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    Why not simply calculating it carefully? It's not allowed to give the full solution in this forum, but here's a hint:

    I'd use
    [tex]\frac{h^2-20h}{(h-10)^2}=\frac{(h-10)^2-100}{(h-10)^2}=1-\frac{100}{(h-10)^2}.[/tex]
    This is trivial to integrate. Mathematica 9.0 under linux gets the correct result. So I wonder, why Wolfram alpha gets it wrong.
     
  10. Mar 17, 2014 #9
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