Đáp án:

• Tìm A và B sao cho  

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

\( =\frac{(A + B)x - 3A - B}{(x - 1)(x - 3)} \)

\( \Leftrightarrow \)  \(\begin{cases} 
A + B = 0 \\ 
-3A - B = 1 
\end{cases}
\Leftrightarrow  
\begin{cases} 
A = -\frac{1}{2} \\ 
B = \frac{1}{2} 
\end{cases}
\)

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

\(=  \frac{1}{2} \left( \ln|x - 3| - \ln|x - 1| \right) = \frac{1}{2} \ln \left| \frac{x - 3}{x - 1} \right| + C\)