If int frac{7 x^8+8 x^7}{left(1+x+x^8right)^2 | vedclass

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Question

(A) To find $f(x)$, we differentiate the given options to see which one yields the integrand $\frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2}$.Let $f(x) =

Vedclass · Accepted Answer

$\frac{x^8}{1+x+x^8}$

Vedclass · Answer

(A) To find $f(x)$, we differentiate the given options to see which one yields the integrand $\frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2}$.Let $f(x) = \frac{x^8}{1+x+x^8}$.Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:$f'(x) = \frac{(1+x+x^8)(8x^7) - x^8(1+8x^7)}{(1+x+x^8)^2}$$f'(x) = \frac{8x^7 + 8x^8 + 8x^{15} - x^8 - 8x^{15}}{(1+x+x^8)^2}$$f'(x) = \frac{7x^8 + 8x^7}{(1+x+x^8)^2}$Since the derivative matches the integrand, $f(x) = \frac{x^8}{1+x+x^8}$ is the correct function.

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$, then $f(x)$ is equal to

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