Đáp án:
\[6 - \frac{{13}}{3}.\ln 5\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^2 {\frac{{3{x^2} - 1}}{{{x^2} - 9}}dx} \\
= \int\limits_0^2 {\frac{{3\left( {{x^2} - 9} \right) + 26}}{{\left( {{x^2} - 9} \right)}}dx} \\
= \int\limits_0^2 {\left( {3 + \frac{{26}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right)dx} \\
= \int\limits_0^2 {\left( {3 + \frac{{13}}{3}.\frac{{\left( {x + 3} \right) - \left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right)dx} \\
= \int\limits_0^2 {\left( {3 + \frac{{13}}{3}.\left( {\frac{1}{{x - 3}} - \frac{1}{{x + 3}}} \right)} \right)dx} \\
= \mathop {\left. {3x + \frac{{13}}{3}.\left( {\ln \left| {x - 3} \right| - \ln \left| {x + 3} \right|} \right)} \right|}\nolimits_0^2 \\
= \mathop {\left. {3x + \frac{{13}}{3}.ln\left| {\frac{{x - 3}}{{x + 3}}} \right|} \right|}\nolimits_0^2 \\
= 6 + \frac{{13}}{3}.\left( {\ln \frac{1}{5} - \ln 1} \right)\\
= 6 - \frac{{13}}{3}.\ln 5
\end{array}\)
