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

\frac{x^{3}}{3} - \frac{x^{2}}{2} + x + C

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

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\frac{x^{3}}{3} - \frac{x^{2}}{2} - x + C

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\frac{x^{3}}{3} + \frac{x^{2}}{2} - x + C

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Solution

The correct option is A $\frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C$
$\int \frac{x^{4} + x^{2} + 1}{x^{2} - x + 1} d x$

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

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

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

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

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

Option B is correct.

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