Đáp án:
\[I = 2\ln 5 - 3\ln 3\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^2 {\frac{{x - 1}}{{{x^2} + 4x + 3}}dx} \\
= \int\limits_0^2 {\frac{{x - 1}}{{\left( {x + 1} \right)\left( {x + 3} \right)}}dx} \\
= \int\limits_0^2 {\frac{{2\left( {x + 1} \right) - \left( {x + 3} \right)}}{{\left( {x + 1} \right)\left( {x + 3} \right)}}dx} \\
= \int\limits_0^2 {\left( {\frac{2}{{x + 3}} - \frac{1}{{x + 1}}} \right)dx} \\
= \mathop {\left. {\left( {2\ln \left| {x + 3} \right| - \ln \left| {x + 1} \right|} \right)} \right|}\nolimits_0^2 \\
= \left( {2\ln 5 - \ln 3} \right) - \left( {2\ln 3 - \ln 1} \right)\\
= 2\ln 5 - 3\ln 3
\end{array}\)
