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

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

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$\frac{1}{4}$

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(C) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 2x^7 + 1)^2} dx$. Divide the numerator and denominator by $x^{14}$: $f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + 2 + x^{-7})^2} dx$. 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: $f(x) = -\int \frac{1}{u^2} du = \frac{1}{u} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$. Since $f(0) = 0$,we evaluate the limit as $x 	o 0^+$. As $x 	o 0^+$,$x^{-5} + x^{-7} 	o \infty$,so $\frac{1}{x^{-5} + x^{-7} + 2} 	o 0$. Thus,$0 = 0 + C$,which gives $C = 0$. Therefore,$f(x) = \frac{1}{x^{-5} + x^{-7} + 2} = \frac{x^7}{1 + x^2 + 2x^7}$. Evaluating at $x = 1$: $f(1) = \frac{1^7}{1 + 1^2 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

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

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