Đáp án:
\[a = 2;\,\,\,b = 39;\,\,c = 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
t = \ln x \Rightarrow \left\{ \begin{array}{l}
dt = \left( {\ln x} \right)'dx = \frac{{dx}}{x}\\
x = {e^2} \Rightarrow t = 2\\
x = {e^3} \Rightarrow t = 3
\end{array} \right.\\
\Rightarrow \int\limits_{{e^2}}^{{e^3}} {\frac{{{{\left( {3\ln x - 2} \right)}^2}}}{{x\left( {\ln x - 1} \right)}}dx} = \int\limits_{{e^2}}^{{e^3}} {\left[ {\frac{{{{\left( {3\ln x - 2} \right)}^2}}}{{\left( {\ln x - 1} \right)}}.\frac{{dx}}{x}} \right]} \\
= \int\limits_2^3 {\frac{{{{\left( {3t - 2} \right)}^2}}}{{t - 1}}dt} \\
= \int\limits_2^3 {\frac{{9{t^2} - 12t + 4}}{{t - 1}}dt} \\
= \int\limits_2^3 {\frac{{9t\left( {t - 1} \right) - 3\left( {t - 1} \right) + 1}}{{t - 1}}dt} \\
= \int\limits_2^3 {\left( {9t - 3 + \frac{1}{{t - 1}}} \right)dt} \\
= \mathop {\left. {\frac{{9{t^2}}}{2} - 3t + \ln \left| {t - 1} \right|} \right|}\nolimits_2^3 \\
= \frac{{39}}{2} + \ln 2 - ln1\\
= \ln 2 + \frac{{39}}{2}\\
\Rightarrow a = 2;\,\,\,b = 39;\,\,c = 2
\end{array}\)
