Đáp án:

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

• \( \int \frac{1}{e^{2x} - 4e^x + 3} \, dx = \int \frac{1}{t (t^2 - 4t + 3)} \, dt \)

• \( \frac{1}{4(t^2 - 4t + 3)} = \frac{1}{t(t-1)(t-3)} = \frac{A}{t} + \frac{B}{t-1} + \frac{C}{t-3} \)

\( \Rightarrow
\begin{cases}
A = \frac{1}{3} \\
B = -\frac{1}{2} \\
C = \frac{3}{2}
\end{cases} \)

• \( I = \int \left( \frac{1}{3t} - \frac{1}{2(t-1)} + \frac{3}{2(t-3)} \right) dt \)

   \( = \frac{1}{3} \ln|t| - \frac{1}{2} \ln|t-1| + \frac{3}{2} \ln|t-3| + c \)

   \( = \frac{1}{3} \ln|e^x| - \frac{1}{2} \ln|e^x - 1| + \frac{3}{2} \ln|e^x - 3| + c \)