If f(x) = int frac{5x^{8} + 7x^{6}}{(x^{2} + | vedclass

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(C) Given $f(x) = \int \frac{5x^{8} + 7x^{6}}{(x^{2} + 1 + 2x^{7})^{2}} dx$. Divide the numerator and denominator by $x^{14}$ inside the integral: $f(

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$4$

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(C) Given $f(x) = \int \frac{5x^{8} + 7x^{6}}{(x^{2} + 1 + 2x^{7})^{2}} dx$. Divide the numerator and denominator by $x^{14}$ inside the integral: $f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^{2}} dx$. Let $t = x^{-5} + x^{-7} + 2$. Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$. Substituting this into the integral: $f(x) = \int -\frac{dt}{t^{2}} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C = \frac{x^{7}}{1 + x^{2} + 2x^{7}} + C$. Given $f(0) = 0$,we find $C = 0$. Thus,$f(x) = \frac{x^{7}}{x^{2} + 1 + 2x^{7}}$. Evaluating at $x = 1$: $f(1) = \frac{1^{7}}{1^{2} + 1 + 2(1)^{7}} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$. Since $f(1) = \frac{1}{K}$,we have $K = 4$.

If $f(x) = \int \frac{5x^{8} + 7x^{6}}{(x^{2} + 1 + 2x^{7})^{2}} dx$,$(x \geq 0)$,$f(0) = 0$ and $f(1) = \frac{1}{K}$,then the value of $K$ is

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