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. $\endgroup$– Parcly TaxelSep 10, 2016 at 11:20
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1 Answer
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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.
