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 perform polynomial long division. Dividing $x^8-9 x^2+18$ by $x^4-3 x^

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 perform polynomial long division. Dividing $x^8-9 x^2+18$ 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 integrand simplifies to $x^4+3 x^2+6$. Now, we integrate the resulting polynomial: $\int (x^4+3 x^2+6) d x = \frac{x^5}{5} + 3 \cdot \frac{x^3}{3} + 6x + c$. Simplifying the expression, we get: $\frac{x^5}{5} + x^3 + 6x + c$.

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

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