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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.2011-07-27
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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.2011-07-27
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    Sorry, I usually post W|A links cause they provide additional useful information. It seems redundant to then additionally include the equation in the post but if you insist I will start writing them out.2011-07-27
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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.2011-07-27
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    I've created a meta post for discussing how self-contained questions should be: http://meta.math.stackexchange.com/questions/2674/how-self-contained-should-questions-be. @mwmnj, I'd appreciate your input there.2011-07-27
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    @J.M. Thanks. I've always wondered what the root-sum thingy meant.2011-07-31
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    @Lyrebird: `Normal[]` *usually* suffices to change `RootSum[]` objects into more familiar expressions, FYI.2011-07-31
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    It may be of interest to note that every rational function has an antiderivative that is an elementary function. Most calculus texts (at least, those from the 1970s and earlier) implicitly show this by covering partial fractions and by showing how to integrate functions of the form $\frac{A}{(ax+b)^n}$ and $\frac{Ax+B}{(ax^2 + bx + c)^n},$ where $n$ is a positive integer and $ax^2 + bx + c$ is an irreducible quadratic. (The irreducible quadratic version for a general positive integer $n$ is usually stated in a recursion format in an exercise or in a table of integrals at the back of the book.)2014-04-25

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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!2011-07-27
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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.)2011-07-27