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What will be the integral of $\dfrac{x^3+x+1}{x^4+x^2+1}$? I tried dividing the denominator by the numerator but it didn't work. I also tried partial fractions but I seem to be doing something wrong.

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  • $\begingroup$ I tried separating all the terms but it becomes quite lengthy. $\endgroup$ Commented May 8, 2021 at 12:39
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    $\begingroup$ @AshthaLohia You need to be more specific about your difficulties... What went wrong with partial fractions? $\endgroup$ Commented May 8, 2021 at 12:40
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    $\begingroup$ hint: $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. $\endgroup$ Commented May 8, 2021 at 12:44

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First note that $x^4+x^2+1 = (x^4+2x^2+1) - x^2 = (x^2+1)^2-x^2 = (x^2+x+1)(x^2-x+1)$ So we have the following: $(x^3+x+1)/(x^4+x^2+1) = ((x+1)(x^2-x+1) + x)/((x^2+x+1)(x^2-x+1))$ The rest should be too simple, try it.

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