Answer
$$\ln \left( {\frac{9}{8}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{x}{{\left( {x - 2} \right)\left( {x - 4} \right)}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{x}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x - 4}} \cr
& x = A\left( {x - 4} \right) + B\left( {x - 2} \right) \cr
& {\text{For }}x = 2 \cr
& 2 = A\left( {2 - 4} \right) \to A = - 1 \cr
& {\text{For }}x = 4 \cr
& 4 = B\left( {4 - 2} \right) \to B = 2 \cr
& \frac{x}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{{ - 1}}{{x - 2}} + \frac{2}{{x - 4}} \cr
& \int_0^1 {\frac{x}{{\left( {x - 2} \right)\left( {x - 4} \right)}}} dx = \int_0^1 {\left( {\frac{{ - 1}}{{x - 2}} + \frac{2}{{x - 4}}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& {\text{ = }}\left[ { - \ln \left| {x - 2} \right| + 2\ln \left| {x - 4} \right|} \right]_0^1 \cr
& {\text{ = }}\left[ { - \ln \left| {1 - 2} \right| + 2\ln \left| {1 - 4} \right|} \right] - \left[ { - \ln \left| {0 - 2} \right| + 2\ln \left| {0 - 4} \right|} \right] \cr
& = \left( {0 + 2\ln 3} \right) - \left( { - \ln 2 + 2\ln 4} \right) \cr
& = 2\ln 3 + \ln 2 - 2\ln 4 \cr
& = \ln 9 + \ln 2 - \ln 16 \cr
& = \ln 18 - \ln 16 \cr
& = \ln \left( {\frac{{18}}{{16}}} \right) \cr
& = \ln \left( {\frac{9}{8}} \right) \cr} $$
