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

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$\frac{x^{3}}{3} + \frac{x^{2}}{2} + x + c$

$\frac{x^{3}}{3} - \frac{x^{2}}{2} + x + c$

$\frac{x^{4}}{4} - \frac{x^{2}}{3} + c$

$\frac{x^{5}}{5} + \frac{x^{3}}{3} + x + c$

answer is D.

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Detailed Solution

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

$= \int \frac{(x^{2} + x + 1) (x^{2} - x + 1)}{(x^{2} - x + 1)} d x$

$= \int (x^{2} + x + 1) d x$

$= \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C$

$n o t e : x^{4} + x^{2} + 1 = x^{4} + x^{2} + 1 + x^{2} - x^{2}$

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

$= (x^{2} + 1 + x) (x^{2} + 1 - x)$

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