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

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(B) 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 \f

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$\frac{1}{4}$

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(B) 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:$f(x) = -\int \frac{dt}{t^2} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.Simplifying the expression:$f(x) = \frac{x^7}{1 + x^2 + 2x^7} + C$.Given $f(0) = 0$,we have $0 = \frac{0}{1} + C$,so $C = 0$.Thus,$f(x) = \frac{x^7}{x^2 + 1 + 2x^7}$.Evaluating at $x = 1$:$f(1) = \frac{1^7}{1^2 + 1 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

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

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