If f(x) = int frac{5x^8 + 7x^6}{(x^2 + 1 + 2x | vedclass

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

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$\frac{1}{4}$

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

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

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