[tex]\displaystyle \int \dfrac{1}{x^2 -3x +2} dx\\\\\\=\displaystyle \int \dfrac{dx}{x^2 -3\cdot \dfrac 32 \cdot x + \left(\dfrac 32 \right)^2 -\left( \dfrac 32 \right)^2 +2}\\\\\\=\displaystyle \int \dfrac{dx}{\left(x- \dfrac 32 \right)^2 +2- \dfrac 94}\\\\\\=\displaystyle \int \dfrac{dx}{\left(x - \dfrac 32 \right)^2-\dfrac 14 }\\\\\\=\displaystyle \int \dfrac{du}{u^2-\left( \dfrac 12 \right)^2} ~~~~~~~~~~~~~~~~~~;\left[u = x - \dfrac32\implies du= dx\right]\\[/tex]
[tex]=\dfrac{1}{2\cdot\tfrac 12} \ln \left| \dfrac{u-\dfrac 12 }{u+\dfrac 12 } \right|+C~~~~~~~~~~:\left[ \displaystyle \int \dfrac {dx}{x^2 -a^2} = \dfrac 1{2a} \ln \left| \dfrac{x-a}{x+a} \right| +C\right]\\\\\\=\ln\left|\dfrac{x- \dfrac 32 - \dfrac 12}{x-\dfrac3 2 + \dfrac 12 } \right|+C~~~~~~~~~~~~;\left[\text{Substitute back}~u =x- \dfrac 32 \right]\\\\\\=\ln\left|\dfrac{x-2}{x-1} \right|+C[/tex]
