int frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x= | vedclass

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(C) To solve the integral $\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x$,we first observe that the degree of the numerator $(8)$ is greater than the degr

Vedclass · Accepted Answer

$\frac{x^5}{5}+x^3+6 x+c$

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(C) To solve the integral $\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x$,we first observe that the degree of the numerator $(8)$ is greater than the degree of the denominator $(4)$.Performing polynomial long division of $x^8-9 x^2+18$ by $x^4-3 x^2+3$:$x^8-9 x^2+18 = (x^4-3 x^2+3)(x^4+3 x^2+6) + 0$.Thus,the integral becomes:$\int (x^4+3 x^2+6) d x$.Integrating term by term with respect to $x$:$= \int x^4 d x + 3 \int x^2 d x + 6 \int 1 d x$.$= \frac{x^5}{5} + 3 \left( \frac{x^3}{3} \right) + 6x + c$.$= \frac{x^5}{5} + x^3 + 6x + c$.

$\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=$

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