If f(x) = int frac{16x^7 + 5x^{10}}{(x^3 + 2 | vedclass

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(A) Given $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$. Divide the numerator and denominator by $x^{16}$: $f(x) = \int \frac{16x^{-9} +

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$\frac{7}{6}$

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(A) Given $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$. Divide the numerator and denominator by $x^{16}$: $f(x) = \int \frac{16x^{-9} + 5x^{-6}}{(x^{-5} + 2x^{-8} + 3)^2} dx$. Let $u = x^{-5} + 2x^{-8} + 3$. Then $du = (-5x^{-6} - 16x^{-9}) dx$, which implies $-du = (16x^{-9} + 5x^{-6}) dx$. Substituting this into the integral: $f(x) = \int -u^{-2} du = u^{-1} + C = \frac{1}{x^{-5} + 2x^{-8} + 3} + C = \frac{x^8}{1 + 2x^5 + 3x^8} + C$. Given $f(0) = 1$, we have $1 = \frac{0}{1} + C$, so $C = 1$. Thus, $f(x) = \frac{x^8}{1 + 2x^5 + 3x^8} + 1$. For $f(1)$, $f(1) = \frac{1}{1 + 2 + 3} + 1 = \frac{1}{6} + 1 = \frac{7}{6}$.

If $f(x) = \int \frac{16x^7 + 5x^{10}}{(x^3 + 2 + 3x^8)^2} dx$ for $x \geq 0$ and $f(0) = 1$, then the value of $f(1)$ is

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