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(C) Given the equation: $\frac{f(x)}{1+x^2} = 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Let $g(x) = \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Then $g'(x) =

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$2e$

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(C) Given the equation: $\frac{f(x)}{1+x^2} = 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Let $g(x) = \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Then $g'(x) = \frac{f(x)}{1+x^2}$. The equation becomes $\frac{f(x)}{1+x^2} = 1 + g(x)$. Substituting $f(x) = (1+x^2)g'(x)$ into the equation: $g'(x) = 1 + g(x)$. This is a linear differential equation: $\frac{dg}{dx} - g = 1$. The integrating factor is $e^{\int -1 dx} = e^{-x}$. Multiplying by $e^{-x}$: $\frac{d}{dx}(g(x)e^{-x}) = e^{-x}$. Integrating both sides: $g(x)e^{-x} = -e^{-x} + C$. So,$g(x) = C e^x - 1$. Since $g(0) = \int_{0}^{0} \dots = 0$,we have $0 = C - 1$,so $C = 1$. Thus,$g(x) = e^x - 1$. Then $g'(x) = e^x$. Since $f(x) = (1+x^2)g'(x)$,we have $f(x) = (1+x^2)e^x$. Therefore,$f(1) = (1+1^2)e^1 = 2e$.

Let $f$ be a differentiable function $f : R \rightarrow R$ satisfying the equation $f(x) = (1+x^2) \left[ 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt \right]$ for all $x \in R$. Then $f(1)$ is:

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