Câu hỏi 37

\(\begin{array}{l}\int\limits_1^0 {\frac{{3x + 1}}{{{x^2} + 2x + 1}}dx}  = \int\limits_1^0 {\frac{{3x + 1}}{{{{\left( {x + 1} \right)}^2}}}dx}  = \int\limits_1^0 {\frac{{3x + 3 - 2}}{{{{\left( {x + 1} \right)}^2}}}dx}  = \int\limits_1^0 {\frac{3}{{x + 1}}dx}  - 2\int\limits_1^0 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}} \\ = \left. {\left( {3\ln \left| {x + 1} \right| + \frac{2}{{x + 1}}} \right)} \right|_1^0 = 2 - 3\ln 2 - 1 = 1 - 3\ln 2\end{array}\)

Chọn D.