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\(\int \frac{x^4+x^2+1}{x^2-x+1} d x=\)
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Verified Answer
The correct answer is:
\(\left(\frac{1}{3}\right) x^3+\left(\frac{1}{2}\right) x^2+x+c\)
We have,
\(\begin{aligned}
& I=\int \frac{x^4+x^2+1}{x^2-x+1} d x \\
& I=\int \frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2-x+1} d x
\end{aligned}\)
So, \(I=\int\left(x^2+x+1\right) d x=\frac{x^3}{3}+\frac{x^2}{2}+x+c\)
\(\begin{aligned}
& I=\int \frac{x^4+x^2+1}{x^2-x+1} d x \\
& I=\int \frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2-x+1} d x
\end{aligned}\)
So, \(I=\int\left(x^2+x+1\right) d x=\frac{x^3}{3}+\frac{x^2}{2}+x+c\)
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