Câu hỏi 16

Ta có \(f\left( x \right) = \dfrac{{3x - 2}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{3x - 6 + 4}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{3}{{x - 2}} + \dfrac{4}{{{{\left( {x - 2} \right)}^2}}}\).

\(\begin{array}{l} \Rightarrow \int {f\left( x \right)dx}  = \int {\left[ {\dfrac{3}{{x - 2}} + \dfrac{4}{{{{\left( {x - 2} \right)}^2}}}} \right]dx} \\ = 3\int\limits_{}^{} {\dfrac{{dx}}{{x - 2}}}  + 4\int\limits_{}^{} {\dfrac{{dx}}{{{{\left( {x - 2} \right)}^2}}}}  = 3\ln \left| {x - 2} \right| - \dfrac{4}{{x - 2}} + C\\Do\,\,x \in \left( {2; + \infty } \right) \Rightarrow x - 2 > 0 \Rightarrow \int {f\left( x \right)dx}  = 3\ln \left( {x - 2} \right) - \dfrac{4}{{x - 2}} + C\end{array}\)

Chọn D