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Evaluate: $$\int\frac{x}{x^3-x^2+1}dx$$

None of the regular methods to integrate rational functions (substitution, partial fractions, etc.) seems to work.

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  • $\begingroup$ The denominator is reducible over the reals, because it is cubic. The one real root is just very complicated to express; for simplicity you can denote it with a letter. $\endgroup$ Sep 10, 2016 at 11:20

1 Answer 1

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The denominator has only one real root, $x_0$. Hence,

$$x^3-x^2+1 = (x-x_0)(x^2+x(x_0-1)+x_0^2-x_0).$$

Then,

$$\frac{x}{x^3-x^2+1} = \frac{A}{x-x_0}+\frac{Bx+C}{x^2+(x_0-1)x+x_0^2-x_0} ,$$

which can be solved for $A,B,C$:

$$A=\frac{1}{3x_0-2},$$ $$B=-A,$$ $$C=A(x_0-1),$$

making the integral approachable.

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