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Integral with partial fractions

  1. Feb 27, 2015 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

    I'm currently in Calculus 3, and the professor gave us a "retro assignment" which is basically a bunch of tough integrals from Calculus 2. I think my process here is valid, but when I check my answer on Wolfram, they're getting a slightly different final answer.

    $$\int\frac{x+6}{x^2(x-3)}dx$$


    3. The attempt at a solution

    Upon first examination, two possibilities popped out at me.

    I could rewrite the function as a sum of two fractions

    $$\int[\frac{x}{x^2(x-3)}+\frac{6}{x^2(x-3)}]dx$$
    $$\int[\frac{1}{x(x-3)}+\frac{6}{x^2(x-3)}]dx$$

    This works, but also involves two partial fraction decompositions. The second possibility that popped out at me was to rewrite the integral in a way that avoided one of the partial fraction decompositions.

    $$\int\frac{x+6}{x^2(x-3)}dx$$
    $$\int\frac{x-3+9}{x^2(x-3)}dx$$
    $$\int[\frac{x-3}{x^2(x-3)}+\frac{9}{x^2(x-3)}]dx$$
    $$\int[\frac{1}{x^2}+\frac{9}{x^2(x-3)}]dx$$
    $$-\frac{1}{x}+9\int\frac{1}{x^2(x-3)}dx$$

    Now I've eliminated one of the partial fraction decompositions. I still have to do a decomposition on the second term.

    $$\frac{1}{x^2(x-3)}$$
    $$\frac{1}{x^2(x-3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x-3)}$$

    Then I multiply by the common denominator

    $$1=Ax(x-3)+B(x-3)+Cx^2$$

    I can see right away that by letting ##x=0## I get

    $$1=-3B \Rightarrow B=-\frac{1}{3}$$

    Similarly, if I let ##x=3##, I get

    $$1=9C \Rightarrow C=\frac{1}{9}$$

    That gives me two of my constants, but there's no immediately apparent value to find ##A##, so I rewrite the decomposition as

    $$1=Ax(x-3)-\frac{1}{3}(x-3)+\frac{1}{9}x^2$$
    $$9=9Ax(x-3)-3(x-3)+x^2$$

    Then I let ##x## be an arbitrary value of 1

    $$9=9A(-2)-3(-2)+1$$
    $$A=-\frac{1}{9}$$

    Now I can rewrite my integral as

    $$-\frac{1}{x}+9\int[\frac{C}{x-3}+\frac{A}{x}+\frac{B}{x^2}]dx$$
    $$-\frac{1}{x}+9\int[\frac{\frac{1}{9}}{x-3}-\frac{\frac{1}{9}}{x}-\frac{\frac{1}{3}}{x^2}]dx$$
    $$-\frac{1}{x}+\int[\frac{1}{x-3}-\frac{1}{x}-\frac{3}{x^2}]dx$$
    $$-\frac{1}{x}+\ln(x-3)-\ln(x)+\frac{3}{x}+C$$

    Which then simplifies to

    $$\frac{2}{x}+\ln(\frac{x-3}{x})+C$$

    Now, when I check my answer in Wolfram, it gives me a solution of

    $$\frac{2}{x}+\ln(3-x)-\ln(x)+C$$

    Which reduces to

    $$\frac{2}{x}+\ln(\frac{3-x}{x})+C$$

    This solution has ##3-x## inside of the logarithm instead of the ##x-3## that I obtained. Did I introduce some kind of error in the way I initially broke the function of by subtracting 3 and adding 9? I've looked it over several times now and I can't find any arithmetic mistakes anywhere. Any help would be very much appreciated. :)
     
  2. jcsd
  3. Feb 27, 2015 #2
    I'm pretty sure you lost the sign on your A.
     
  4. Feb 27, 2015 #3

    QuantumCurt

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    Are you sure? I just pulled it out in front of the whole fraction when I substituted the numerical values in.

    $$-\frac{1}{x}+9\int[\frac{C}{x-3}+\frac{A}{x}+\frac{B}{x^2}]dx$$
    $$-\frac{1}{x}+9\int[\frac{\frac{1}{9}}{x-3}-\frac{\frac{1}{9}}{x}-\frac{\frac{1}{3}}{x^2}]dx$$
     
  5. Feb 27, 2015 #4

    Mark44

    Staff: Mentor

    @QuantumCurt, I haven't checked your work... When you use decomposition by partial fractions, it's a good idea to check your work by adding the (in this case, 3) fractions to make sure you end up with what you started.
     
  6. Feb 27, 2015 #5
    Sorry, I missed that you switched the order of A and C.
     
  7. Feb 27, 2015 #6

    QuantumCurt

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    @Mark44 , that's good advice. I hadn't added the decomposition back together to check that it came out the same. I just checked it, and it does indeed come out the same.
     
  8. Feb 27, 2015 #7

    Mark44

    Staff: Mentor

    OK, then since the numbers check, you can check your integration by taking the derivative of your answer. With a bit of albebraic manipulation, you should be able to get back to the original integrand.
     
  9. Feb 27, 2015 #8

    QuantumCurt

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    Also good advice. I just took the derivative of my answer, and after combining the fractions I ended up with my original function.

    So...am I right in thinking that Wolfram might simply be getting this one wrong?
     
  10. Feb 27, 2015 #9

    Dick

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    You are both somewhat right. The derivative of log(3-x) is the same as the derivative of log(x-3). The usual thing is to write the integral of 1/(x-3) as log(|x-3|).
     
  11. Feb 27, 2015 #10

    QuantumCurt

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    Now that you mention that...I recall encountering this issue with Wolfram before when functions have integrated to logarithms. It often seems to put it in the log(c-x) form rather than log(x-c).
     
  12. Feb 27, 2015 #11

    QuantumCurt

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    Thanks for the help everyone!

    I just finished my retro calculus assignment, and that was the only one that I got hung up on. Good to know that I still remember all of this stuff from calculus 2.
     
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