Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
62,\\
\int\limits_0^2 {\frac{{x - 1}}{{{x^2} + 4x + 3}}dx} \\
= \int\limits_0^2 {\frac{{2\left( {x + 1} \right) - \left( {x + 3} \right)}}{{\left( {x + 1} \right)\left( {x + 3} \right)}}dx} \\
= \int\limits_0^2 {\left( {\frac{2}{{x + 3}} - \frac{1}{{x + 1}}} \right)dx} \\
= 2\int\limits_0^2 {\frac{{d\left( {x + 3} \right)}}{{x + 3}}} - \int\limits_0^2 {\frac{{d\left( {x + 1} \right)}}{{x + 1}}} \\
= \mathop {2\ln \left| {x + 3} \right| - \ln \left| {x + 1} \right|}\nolimits_0^2 \\
= 2.\ln 5 - 2\ln 3 - \ln 3 + \ln 1\\
= 2\ln 5 - 3\ln 3\\
63,\\
\int\limits_0^1 {\frac{{4x + 11}}{{{x^2} + 5x + 6}}dx} \\
= \int\limits_0^1 {\frac{{\left( {x + 2} \right) + 3\left( {x + 3} \right)}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}dx} \\
= \int\limits_0^1 {\left( {\frac{1}{{x + 3}} + \frac{3}{{x + 2}}} \right)dx} \\
= \int\limits_0^1 {\frac{{d\left( {x + 3} \right)}}{{x + 3}}} + 3\int\limits_0^1 {\frac{{d\left( {x + 2} \right)}}{{x + 2}}} \\
= \mathop {\ln \left| {x + 3} \right| + 3\ln \left| {x + 2} \right|}\nolimits_0^1 \\
= \ln 4 - \ln 3 + 3\ln 3 - 3\ln 2\\
= 2\ln 2 + 2\ln 3 - 3\ln 2\\
= \ln 9 - \ln 2 = \ln \frac{9}{2}
\end{array}\)
