Câu hỏi 16

Ta có:

\(\eqalign{
& \int\limits_1^3 {{{2x + 1} \over {{x^2} + 3x + 2}}dx} = \int\limits_1^3 {{{2x + 1} \over {\left( {x + 1} \right)\left( {x + 2} \right)}}dx} \cr
& = \int\limits_1^3 {\left( {{3 \over {x + 2}} - {1 \over {x + 1}}} \right)dx} = \left. {\left( {3\ln \left| {x + 2} \right| - \ln \left| {x + 1} \right|} \right)} \right|_1^3 \cr
& = 3\ln 5 - \ln 4 - 3\ln 3 + \ln 2 = 3\ln 5 - 2\ln 2 - 3\ln 3 + \ln 2 \cr
& = - \ln 2 - 3\ln 3 + 3\ln 5 = a\ln 2 + b\ln 3 + c\ln 5 \cr
& \Rightarrow \left\{ \matrix{
a = - 1 \hfill \cr
b = - 3 \hfill \cr
c = 3 \hfill \cr} \right. \Rightarrow a + b + c = - 1 - 3 + 3 = - 1. \cr} \)

Chọn  A.