Answer
$$x = \ln \left| {\frac{{t - 2}}{{t - 1}}} \right| + \ln 2$$
Work Step by Step
$$\eqalign{
& \left( {{t^2} - 3t + 2} \right)\frac{{dx}}{{dt}} = 1 \cr
& {\text{Separate the variables}} \cr
& dx = \frac{{dt}}{{{t^2} - 3t + 2}} \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {dx} = \int {\frac{{dt}}{{{t^2} - 3t + 2}}} \cr
& x = \int {\frac{{dt}}{{{t^2} - 3t + 2}}} \cr
& {\text{Decompose }}\frac{1}{{{t^2} - 3t + 2}}{\text{ into partial fractions}} \cr
& \frac{1}{{{t^2} - 3t + 2}} = \frac{1}{{\left( {t - 2} \right)\left( {t - 1} \right)}} \cr
& \frac{1}{{\left( {t - 2} \right)\left( {t - 1} \right)}} = \frac{A}{{t - 2}} + \frac{B}{{t - 1}} \cr
& {\text{Multiply by }}\left( {t - 2} \right)\left( {t - 1} \right){\text{ and simplify}} \cr
& 1 = A\left( {t - 1} \right) + B\left( {t - 2} \right) \cr
& {\text{if }}t = 2,\,\,\,A = 1 \cr
& {\text{if }}t = 1,\,\,\,B = - 1 \cr
& \frac{1}{{\left( {t - 2} \right)\left( {t - 1} \right)}} = \frac{1}{{t - 2}} - \frac{1}{{t - 1}} \cr
& \cr
& x = \int {\left( {\frac{1}{{t - 2}} - \frac{1}{{t - 1}}} \right)} dt \cr
& {\text{Integrate}} \cr
& x = \ln \left| {\frac{{t - 2}}{{t - 1}}} \right| + C \cr
& {\text{Use }}x\left( 3 \right) = 0 \cr
& 0 = \ln \left| {\frac{{3 - 2}}{{3 - 1}}} \right| + C \cr
& C = -\ln \frac{1}{2} \cr
& C = \ln 2 \cr
& \cr
& {\text{Then}} \cr
& x = \ln \left| {\frac{{t - 2}}{{t - 1}}} \right| + \ln 2 \cr} $$
