\(\int\dfrac{1}{x^2-4x+3}dx=\int\dfrac{1}{\left(x-1\right)\left(x-3\right)}dx=\dfrac{1}{2}\int\dfrac{2}{\left(x-1\right)\left(x-3\right)}dx\)
\(=\dfrac{1}{2}\int\left(\dfrac{1}{x-3}-\dfrac{1}{x-1}\right)dx\)
\(=\dfrac{1}{2}\left(ln\left|x-3\right|-ln\left|x-1\right|\right)+C\)
\(=\dfrac{1}{2}ln\left|\dfrac{x-3}{x-1}\right|+C\)