Đáp án:

\(\int_{0}^{1} \left[\frac{f'(x)}{e^{\frac{x}{2}}}\right]^2 dx = \frac{1}{e - 1}.\)

Đặt \(h(x) = \frac{f'(x)}{e^{\frac{x}{2}}} \Rightarrow f'(x) = e^{\frac{x}{2}} h(x)\).

Tìm \(k\) để: \(\int_{0}^{1} \left[h(x) - k e^{\frac{x}{2}}\right]^2 dx = 0.\)

\(\Leftrightarrow \int_{0}^{1} [h(x)]^2 dx - 2k \int_{0}^{1} f'(x)  dx + k^2 \int_{0}^{1} e^x dx = 0.\)

\(\Leftrightarrow \frac{1}{e - 1} - 2k f(x) \big|_{0}^{1} + k^2 e^x \big|_{0}^{1} = 0.\)

\(\Leftrightarrow (e - 1)k^2 - 2k + \frac{1}{e - 1} = 0 \quad \Leftrightarrow \quad k = \frac{1}{e - 1}\)
\(h(x) = \frac{e^{\frac{x}{2}}}{e - 1} \Rightarrow \frac{f'(x)}{e^{\frac{x}{2}}} = \frac{e^{\frac{x}{2}}}{e - 1}\)

\(f'(x) = \frac{e^{x}}{e - 1} \quad \Rightarrow \quad f(x) =  \frac{e^{x}}{e - 1} + C.\)

\(f(0) = 0 \quad \Rightarrow \quad C = -\frac{1}{e - 1}.\)

\(\Rightarrow f(x) = \frac{e^{x}}{e - 1} - \frac{1}{e - 1} \Rightarrow f(x) = \frac{e^{\frac{x}{2}} - 1}{e - 1}.\)

\(\int_{0}^{1} f(x) dx = \int_{0}^{1} \frac{e^{x} - 1}{e - 1} dx = \frac{e - 2}{e - 1} \Rightarrow \boxed{A}.\)