Đáp án:
\(B\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\int\limits_0^1 {\dfrac{{xdx}}{{{{\left( {x + 2} \right)}^2}}}} \\
= \int\limits_0^1 {\dfrac{{x + 2 - 2}}{{{{\left( {x + 2} \right)}^2}}}dx} \\
= \int\limits_0^1 {\left[ {\dfrac{1}{{x + 2}} - \dfrac{2}{{{{\left( {x + 2} \right)}^2}}}} \right]dx} \\
= \int\limits_0^1 {\dfrac{{d\left( {x + 2} \right)}}{{x + 2}} - 2\int\limits _0^1{\dfrac{{d\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}}}} } \\
= \mathop {\ln \left| {x + 2} \right||_0^1 + \dfrac{2}{{x + 2}}}\nolimits|_0^1 \\
= \ln 3 - \ln 2 + \dfrac{2}{3} - \dfrac{2}{2} = - \dfrac{1}{3} - \ln 2 + \ln 3\\
\Rightarrow a = - \dfrac{1}{3};\,\,\,\,b = - 1;\,\,\,\,c = 1\\
\Rightarrow 3a + b + c = - 1
\end{array}\)
