If int frac{x^8+4}{x^4-2 x^2+2} d x=A x^5+B x | vedclass

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Question

(B) We are given the integral $\int \frac{x^8+4}{x^4-2 x^2+2} d x$. First,we rewrite the numerator by adding and subtracting $4x^4$: $\int \frac{x^8+4

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) We are given the integral $\int \frac{x^8+4}{x^4-2 x^2+2} d x$. First,we rewrite the numerator by adding and subtracting $4x^4$: $\int \frac{x^8+4x^4+4-4x^4}{x^4-2 x^2+2} d x = \int \frac{(x^4+2)^2 - (2x^2)^2}{x^4-2 x^2+2} d x$. Using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$,we have: $\int \frac{(x^4+2-2x^2)(x^4+2+2x^2)}{x^4-2 x^2+2} d x = \int (x^4+2x^2+2) d x$. Integrating term by term,we get: $\frac{x^5}{5} + \frac{2x^3}{3} + 2x + k$. Comparing this with $Ax^5+Bx^3+Cx+k$,we find $A = \frac{1}{5}$,$B = \frac{2}{3}$,and $C = 2$. Now,calculate $5A+3B+C$: $5(\frac{1}{5}) + 3(\frac{2}{3}) + 2 = 1 + 2 + 2 = 5$.

If $\int \frac{x^8+4}{x^4-2 x^2+2} d x=A x^5+B x^3+C x+k$,then $5 A+3 B+C=$

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