Câu hỏi 20

Ta có:

\(\begin{array}{l}I = \int {\frac{{2x - 1}}{{{{\left( {x + 1} \right)}^2}}}dx}  = \int {\frac{{2\left( {x + 1} \right) - 3}}{{{{\left( {x + 1} \right)}^2}}}dx}  = \int {\frac{2}{{x + 1}}dx - } \int {\frac{3}{{{{\left( {x + 1} \right)}^2}}}dx} \\\,\,\,\, = 2\ln \left| {x + 1} \right| + \frac{3}{{x + 1}} + C = 2\ln \left( {x + 1} \right) + \frac{3}{{x + 1}} + C\,\,\,\,\,\left( {do\,\,x \in \left( { - 1; + \infty } \right) \Rightarrow x + 1 > 0} \right).\end{array}\)

Chọn  B.