Đáp án:

• Đặt \( t = \sqrt{1 + e^x} \Rightarrow e^x = t^2 - 1 \)

• \(\Rightarrow e^x dx = 2t dt \Rightarrow dx = \frac{2t}{t^2 - 1} \, dt \)

• \( \int \frac{e^{2x}}{\sqrt{1 + e^x}} \, dx = \int \frac{\left( t^2 - 1 \right)^2}{t} \cdot \frac{2t}{(t^2 - 1)} \, dt \)

\( = \int 2(t^2 - 1) \, dt = 2 \frac{t^3}{3} - 2t + C \)