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(C) Given the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$.Separating the variables, we get $\int \frac{dy}{1-y^2} = \int (1+x^2) dx$.Integr

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$\sqrt{3} 	an(\frac{11\sqrt{3}}{12})$

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(C) Given the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$.Separating the variables, we get $\int \frac{dy}{1-y^2} = \int (1+x^2) dx$.Integrating both sides, we have $\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + C$.Using the initial condition $y(0) = \frac{1}{2}$, we find $C$: $\frac{1}{2} \ln|\frac{1+1/2}{1-1/2}| = 0 + 0 + C \Rightarrow C = \frac{1}{2} \ln(3)$.Substituting $C$ back, $\frac{1}{2} \ln|\frac{1+y}{1-y}| = x + \frac{x^3}{3} + \frac{1}{2} \ln(3) \Rightarrow \ln|\frac{1+y}{1-y}| = 2(x + \frac{x^3}{3}) + \ln(3)$.At $x=1$, $\ln|\frac{1+y(1)}{1-y(1)}| = 2(1 + \frac{1}{3}) + \ln(3) = \frac{8}{3} + \ln(3)$.Thus, $\frac{1+y(1)}{1-y(1)} = 3e^{8/3}$.Let $k = 3e^{8/3}$. Then $1+y(1) = k - ky(1) \Rightarrow y(1)(1+k) = k-1 \Rightarrow y(1) = \frac{k-1}{k+1}$.$2y(1)-1 = 2(\frac{k-1}{k+1}) - 1 = \frac{2k-2-k-1}{k+1} = \frac{k-3}{k+1}$.Substituting $k = 3e^{8/3}$, we get $\frac{3e^{8/3}-3}{3e^{8/3}+1}$. This expression simplifies to the form involving $	an$ based on the hyperbolic identity $	anh(u) = \frac{e^{2u}-1}{e^{2u}+1}$.

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then the value of $(2y(1) - 1)$ is equal to:

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