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How do I calculate the following indefinite integral?

$$\int\frac{1}{x^3+x+1}dx$$


Approach:

$x^3+x+1=(x-a)(x^2+ax+c)$ where

$a:$ real solution of the equation $a^3+a+1=0$

$c:$ real solution of the equation $c^3-c^2+1=0$

Then $$\int\frac{1}{x^3+x+1}dx=\int\frac{1}{(x-a)(x^2+ax+c)}dx=\int\frac{A}{(x-a)}dx+\int\frac{Bx+C}{(x^2+ax+c)}dx$$

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  • $\begingroup$ The problem is, how are you going to factor the denom nicely? $\endgroup$ – imranfat Mar 29 '16 at 20:09
  • $\begingroup$ The WA solution factored the denominator completely using Cardano's formula, then expanded into linear complex factors and integrated. To keep everything real, you could use Vieta's method to find the real root $a=-\frac2{\sqrt3}\sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)$ $\endgroup$ – user5713492 Mar 29 '16 at 20:22
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Starting with the root from my previous comment, $$a=-\frac2{\sqrt3}\sinh\left(\frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$ We can factor the denominator as $x^3+x+1=(x-a)(x^2+ax+a^2+1)$. Then the partial fractions expansion reads $$\frac1{x^3+x+1}=\frac{A}{x-a}+\frac{Bx+C}{x^2+ax+a^2+1}$$ We can find $A$ by multiplying both sides by $(x-a)$ and taking the limit: $$A=\lim_{x\rightarrow a}\frac{x-a}{x^3+x+1}=\frac1{3a^2+1}$$ by L'Hopital's rule. If we observe that $$\begin{align}\left(3a^2+1\right)\left(6a^2-9a+4\right) & =\left(18a-27\right)\left(a^3+a+1\right)+31 \\ & =31\end{align}$$ It follows that $$A=\frac{6a^2-9a+4}{31}$$ Then if we multiply through by $(x^3+x+1)$ and compare coefficients of like powers of $x$ we find that $$\begin{align}B & =\frac{-6a^2+9a-4}{31} \\ C & =\frac{18a^2+4a+12}{31}\end{align}$$ So $$\begin{align}\frac1{x^3+x+1} & =\frac1{31}\left\{\frac{6a^2-9a+4}{x-a}+\frac{-\left(6a^2-9a+4\right)x+18a^2+4a+12}{x^2+ax+a^2+1}\right\} \\ & =\frac1{31}\left\{\frac{6a^2-9a+4}{x-a}+\frac{\left(-3a^2+\frac92a-2\right)(2x+a)+\frac{27}2a^2+3a+9}{x^2+ax+a^2+1}\right\}\end{align}$$ Now all our integrals are elementary and we find $$\int\frac1{x^3+x+1}dx=\frac1{31}\left\{\left(6a^2-9a+4\right)\ln|x-a|-\left(3a^2-\frac92a+2\right)\ln\left(x^2+ax+a^2+1\right)+\frac{\left(27a^2+6a+18\right)}{\sqrt{3a^2+4}}\tan^{-1}\left(\frac{2x+a}{\sqrt{3a^2+4}}\right)\right\}+C$$ Numerical integration confirms this result.

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    $\begingroup$ +1 for the sheer effort for doing this and writing it all up. :) $\endgroup$ – Deepak Mar 30 '16 at 3:07
  • $\begingroup$ Really nice and elegant answer ! $\endgroup$ – Claude Leibovici Mar 30 '16 at 8:10

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