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I was bored and started punching in some rational functions into WolframAlpha to integrate and I came across a closed form that I've never seen before and have no clue how it would even be derived.

The integral I took was $$ \int{\frac{dx}{Ax^3 + Bx^2 + Cx + D}} $$ and the return value given was $$ \sum_{\{\omega\ : \ A\omega^3 + B\omega^2 + C\omega + D = 0\}}{\frac{\ln{|-\omega + x|}}{3A\omega^2 + 2B\omega + C}} \color{silver}{+ \text{constant}} $$ It seems as though in general, $$ \int{\frac{dx}{\sum_{k=1}^n{A_kx^{n-k}}}} = \sum_{\{\omega\ : \ \sum_{k=1}^n{A_k\omega^{n-k}}=0\}}{\frac{\ln{|-\omega + x|}}{\frac{d}{dx}{\left(\sum_{k=1}^n{A_k\omega^{n-k}}\right)}}} \color{silver}{+ \text{constant}} $$ Where does this come from? Is there a name for this type of indefinite integral?

user3002473
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1 Answers1

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Partial fractions: if $P(x)$ is a polynomial of degree $n$ with $n$ distinct roots $\omega_1, \ldots, \omega_n$, then $$ \dfrac{1}{P(x)} = \sum_{j=1}^n \dfrac{c_j}{x - \omega_j}$$ where $c_j$ is the residue of $1/P(x)$ at $x=\omega_j$, which is $1/P'(\omega_j)$. Then integrate.

Robert Israel
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  • Oh gosh, residues, I have absolutely no knowledge of complex analysis. The time will come when I can fully appreciate this answer, but for now thank you for elaborating! – user3002473 Sep 01 '14 at 03:17
  • @user3002473 You don't need complex analysis. All you need to know is to do partial fractions. (Yes, one way is to use residues, but there are other methods.) – Akiva Weinberger Sep 01 '14 at 03:53
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    Once you know $$ \dfrac{1}{P(x)} = \sum_{j=1}^n \dfrac{c_j}{x-\omega_j}$$ just multiply both sides by $x - \omega_j$ and take the limit as $x \to \omega_j$, using l'Hospital's rule on the left. – Robert Israel Sep 01 '14 at 04:08