int {frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1 | vedclass

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Question

(A) To solve the integral $\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx}$,we first simplify the integrand by factoring the numerator.We know

Vedclass · Accepted Answer

$\frac{1}{3}{x^3} + \frac{1}{2}{x^2} + x + c$

Vedclass · Answer

(A) To solve the integral $\int {\frac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}\,dx}$,we first simplify the integrand by factoring the numerator.We know that ${x^4} + {x^2} + 1 = ({x^2} + 1)^2 - {x^2} = ({x^2} + 1 - x)({x^2} + 1 + x)$.Substituting this into the integral,we get:$\int {\frac{({x^2} - x + 1)({x^2} + x + 1)}{{{x^2} - x + 1}}\,dx} = \int {({x^2} + x + 1)\,dx}$.Now,integrate term by term:$\int {x^2\,dx} + \int {x\,dx} + \int {1\,dx} = \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + c$.Thus,the correct option is $A$.

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

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