int frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x= | vedclass

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Question

(A) To solve the integral $I = \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x$,we perform polynomial long division.Dividing $x^4-16 x^2+2 x+8$ by $x^3-

Vedclass · Accepted Answer

$\frac{x^2+8 x+c}{2}$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x$,we perform polynomial long division.Dividing $x^4-16 x^2+2 x+8$ by $x^3-4 x^2+2$:$x^4-16 x^2+2 x+8 = (x+4)(x^3-4 x^2+2) + (0x^2 + 0x + 0)$.Since the remainder is $0$,the expression simplifies to:$I = \int (x+4) d x$.Integrating term by term:$I = \frac{x^2}{2} + 4x + C = \frac{x^2+8x}{2} + C$.

$\int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x=$

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