If int frac{7x^8+8x^7}{(1+x+x^8)^2} dx = f(x) | vedclass

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Question

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

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

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

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

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