DC2

The given series is geometric with common ratio $r = f'(x)$ , and since we're given that $\text{-}1 < f'(x) < 1$ , we know it's sum is given by the formula $\dfrac{a}{1-r}$ . Thus

$\begin{aligned} e^x &= \dfrac{f'(x)}{1 - f'(x)} \\ \\ f'(x) &= \dfrac{e^x}{1 + e^x} \\ \\ f(x) &= \ln (1 + e^x) + C \\ \\ f(0) = 0 &\implies \ln(2) + C = 0 \implies C = -\ln(2) \\ \\ f(x) = \ln(1 + e^x) - \ln(2) &\implies f(1) = \ln(1 + e) - \ln(2) \approx \boxed{0.620114507}\end{aligned}$