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Homework Help: Integration of Rational Functions by Partial Fractions

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral. (Remember to use ln |u| where appropriate.)

    ∫(x^3 + 36)/(x^2 + 36)

    2. Relevant equations

    3. The attempt at a solution

    A little bit confused about arriving at the solution for this problem. I get stuck a little ways in. Any help would be greatly appreciated!

    First I divided x^3+36 by the denominator, since the power of the numerator is greater. Doing so, I got x + (-36x+36)/(x^2+36). I rewrote my integral as ∫ x + (-36x+36)/(x^2+36).

    Now, I want to be able to use some facet of partial fractions, the whole A + B...etc, but I don't really see how I can in this case. If I split up the second half of the integral, I can write it as -36x/x^2+36 + 36/x^2+6^2. Applying b^2-4ac < 0, I know that I won't be able to factor and that I could complete the square in these cases, but I don't really see how that helps me either.

    I don't think the problem calls for u-substitution, as I've tried that, and since the problem is part of the partial fractions section, I'm thinking they want us to go about using that method.

    I'm not looking for an exact answer, just a little confused about what method to use from here.

    Thanks in advance!
  2. jcsd
  3. Feb 14, 2012 #2


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    You won't be able to, because x2 +36 can't be factored. Break it up as a sum of two fractions:
    [itex]\int x dx + \int \frac{-36x}{x^2 + 36} dx + \int \frac{36}{x^2 + 36} dx[/itex]
    (You might as well factor out a 36 from the 2 fractions before proceeding.)

    You can use a u-substitution from the 1st fraction. The 2nd fraction... needs something else entirely. You familiar with the integrals that involve inverse trig?
  4. Feb 14, 2012 #3
    I am familiar with inverse trig functions for integrals, and so for the second fraction, I see that if I can get it into the form 1/x^2+1, the integral is simply tan^-1.

    So if I factor 36 from each fraction, I get

    ∫x/(x^2/36+1) + ∫1/(x^2/36 +1)

    For the first fraction, u=x/6, du=1/6dx, so I now have ∫36u/(u^2+1) du. Similarly, for the second fraction I get, ∫1/(x^2/36+1). Setting u=x/6 for this fraction as well, and integrating, I get 6tan^-1(x/6).

    Now for the first fraction...I'm stuck. I know the solution entails ln(x^2+36) which doesn't make any sense to me, because I thought that x^2+1 suggests we will use tan^-1, as we did in the second fraction.
  5. Feb 14, 2012 #4


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    The u on top in
    $$36 \int \frac{u}{u^2+1}\,du$$ is the reason you get a log function. You're over-thinking it a bit. As eumyang said, you can do this one with a simple substitution.
  6. Feb 14, 2012 #5
    So essentially the u in the numerator cancels with a u on the bottom? And that's why we get the log function?
  7. Feb 14, 2012 #6


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    If you're saying what I think you're saying, that is,
    $$\frac{u}{u^2+1} \rightarrow \frac{1}{u+1},$$ then no. Algebra doesn't work that way.
  8. Feb 14, 2012 #7
    Haha I know, that's why I'm completely confused as to how u/u^2+1 becomes the log function?
  9. Feb 14, 2012 #8
    Nevermind, I figured it out. It takes another substitution, in this case u^2+1=z.

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