Answer
$4\ln \left| {\sqrt x - 2} \right| - 2\ln \left| {\sqrt x - 1} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x - 3\sqrt x + 2}}} dx \cr
& {\text{Substitute }}x = {u^2},{\text{ then }}dx = 2udu \cr
& = \int {\frac{1}{{{u^2} - 3\sqrt {{u^2}} + 2}}\left( {2u} \right)} du \cr
& = \int {\frac{{2u}}{{{u^2} - 3u + 2}}} du \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{2u}}{{{u^2} - 3u + 2}} = \frac{{2u}}{{\left( {u - 2} \right)\left( {u - 1} \right)}} = \frac{A}{{u - 2}} + \frac{B}{{u - 1}} \cr
& {\text{Multiply both sides by the least common denominator}} \cr
& 2u = A\left( {u - 1} \right) + B\left( {u - 2} \right) \cr
& u = 2 \to A = 4 \cr
& u = 1 \to B = - 2 \cr
& \frac{{2u}}{{\left( {u - 2} \right)\left( {u - 1} \right)}} = \frac{4}{{u - 2}} - \frac{2}{{u - 1}} \cr
& \int {\frac{{2u}}{{{u^2} - 3u + 2}}} du = \int {\left( {\frac{4}{{u - 2}} - \frac{2}{{u - 1}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 4\ln \left| {u - 2} \right| - 2\ln \left| {u - 1} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ }}x = {u^2} \to u = \sqrt x \cr
& = 4\ln \left| {\sqrt x - 2} \right| - 2\ln \left| {\sqrt x - 1} \right| + C \cr} $$
