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I came across an integral problem, it was a solved example, which goes something like this.

$$\int \frac{x^4}{x^2-x+1} dx$$ They straight away factored above integral into $$\int x^2+x - \frac{x}{x^2-x+1}dx$$ and went about solving it but what I found difficult was how did they factored from intital to second form

$$\frac{x^4}{x^2-x+1} \Rightarrow x^2+x+\frac{x}{x^2-x+1}$$

Thanks for your help on this.

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  • $\begingroup$ Polynomial long division, synthetic division, the Euclidan algorithm, take your pick. $\endgroup$ – Doug M Jan 11 '20 at 4:53
  • $\begingroup$ Can you illustrate any one of them with above example. $\endgroup$ – Sid111Math Jan 11 '20 at 4:55
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$\quad\dfrac{x^4}{x^2-x+1} $

$= \dfrac{x^4-x^3+x^2}{x^2-x+1} +\dfrac{x^3-x^2}{x^2-x+1}$

$= \dfrac{x^4-x^3+x^2}{x^2-x+1} +\dfrac{x^3-x^2+x}{x^2-x+1} - \dfrac{x}{x^2-x+1}$

$=\qquad x^2 \qquad\quad+\qquad x \quad\qquad- \dfrac{x}{x^2-x+1}$

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$\frac {x^4}{x^2 - x +1} = \frac {x^2(x^2 - x + 1) + x^3 - x^2}{x^2 - x + 1} = \frac {x^2(x^2 - x + 1) + x(x^2-x + 1) + x^2 - x - x^2}{x^2 - x + 1} = \frac {x^2(x^2 - x + 1) + x(x^2-x + 1) - x }{x^2 - x + 1}$

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Write $\frac 1{x^2-x+1} = \frac{x+1}{x^3+1}$ and $x^4=x(x^3+1-1)$ to decompose

$$\frac{x^4}{x^2-x+1} = \frac{x(x+1)(x^3+1-1)}{x^3+1} = x(x+1)-\frac{x(x+1)}{x^3+1} =x^2+x-\frac x{x^2-x+1}$$

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$$\frac{x^4}{x^2-x+1}=\frac{x^4+x-x}{x^2-x+1}=x^2+x-\frac{x}{x^2-x+1}.$$

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    $\begingroup$ Someone want's to delete my solution. Explain please your step. $\endgroup$ – Michael Rozenberg Jan 11 '20 at 6:36

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