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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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    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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    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