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The integral that I am trying to evaluate is $$\int \frac{7-x}{x^3-x^2-x-2}dx.$$ Here is the Wolfram Alpha link: W|A Link

Thanks to all who take a look!

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@mwmnj: This is the third question you posted today that wasn't self-contained and merely contained a link to W|A. Each of the others was edited to include the question, and on the second one I commented: "Your previous post had also been edited by someone else to make it self-contained. Please try to learn from such edits so they won't be required in the future." If you disagree with the policy of making questions self-contained, please provide arguments so we can find the best approach together. Please don't just ignore the community norms without responding to suggestions. –  joriki Jul 27 '11 at 15:43
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FWIW: the RootSum[] function in Mathematica, as the name might probably imply, means that you take the sum of the function in its second argument over the roots of the polynomial in the argument. Thus, $-\sum_{i=1}^3\frac{(x_i-7)\log(x-x_i)}{3x_i^2-4x_i-1}$ where the $x_i$ satisfy $x_i^3-2x_i^2-x_i-2=0$ is the result. –  J. M. Jul 27 '11 at 15:48
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@mwmnj: The idea is that the question should be self contained. As an answerer, it can be annoying to have to visit another site to even know what the question is. Also, some people don't completely like Wolfram and its encyclopedia math pages. Although I find it to be an incredibly useful tool, there are some pages with incorrect statements. –  Eric Naslund Jul 27 '11 at 15:54
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@mwmnj: What if Wolfram Alpha goes down or is removed from free public use? And if these are homework, you might want to tag them as such and explain some of your efforts. And @ Rest, IMO, a link to Wolfram Alpha is not a good example of showing effort. –  Aryabhata Jul 27 '11 at 16:12
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I've created a meta post for discussing how self-contained questions should be: meta.math.stackexchange.com/questions/2674/…. @mwmnj, I'd appreciate your input there. –  joriki Jul 27 '11 at 16:24
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1 Answer

up vote 3 down vote accepted

The polynomial $x^{3}-2x^{2}-x-2$ does not factor nicely at all, so this problem will have an incredibly ugly and complicated solution as you saw on Wolfram Alpha. There is no way to get around this.

However, I'll assume this is for a calculus course you are taking. Then, I believe there there is a typo and that the denominator should be $x^{3}-2x^{2}-x+2$ (the last sign should be $+$ instead of $-$). This polynomial factors very nicely as $(x-2)(x-1)(x+1).$ Then in this case partial fractions yields $$\frac{7-x}{x^{3}-2x-x+2}=\frac{-3}{x-1}+\frac{4}{3(x+1)}+\frac{5}{3(x-2)},$$

so that

$$\int\frac{7-x}{x^{3}-2x-x+2}dx=-3\ln|x-1|+\frac{4}{3}\ln|x+1|+\frac{5}{3}\ln|x-2|+C.$$

Hope that helps,

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Ill have to check if its a typo, that would make sense thanks! –  Matt Jul 27 '11 at 15:53
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If it doesn't factor "nicely", it still factors a 1st-degree real factor times a quadratic factor, where you might want to use numerical methods to find the factors. Regardless of whether the quadratic factor can be factored using real numbers, you can apply partial fractions to find the integral. (But that probably would not appear without advance warning as an exercise. But maybe the advance warning would be only an oral statement by the instructor in class. Especially if the instructor is naive and thinks students always understand everything.) –  Michael Hardy Jul 27 '11 at 19:04
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