Đáp án:
\[2\ln \left| {x - 3} \right| - \ln \left| {x - 1} \right| - \ln \left| {x - 2} \right|\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
F\left( x \right) = \int {f\left( x \right)dx} \\
= \int {\frac{{3x - 5}}{{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)}}dx} \\
= \int {\frac{{3x - 5}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx} \\
= \int {\frac{{2\left( {{x^2} - 3x + 2} \right) - \left( {{x^2} - 5x + 6} \right) - \left( {{x^2} - 4x + 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx} \\
= \int {\frac{{2\left( {x - 1} \right)\left( {x - 2} \right) - \left( {x - 2} \right)\left( {x - 3} \right) - \left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}dx} \\
= \int {\left( {\frac{2}{{x - 3}} - \frac{1}{{x - 1}} - \frac{1}{{x - 2}}} \right)dx} \\
= 2\ln \left| {x - 3} \right| - \ln \left| {x - 1} \right| - \ln \left| {x - 2} \right|
\end{array}\)
