Đáp án D

Chọn D
Có \(f'\left( x \right) = \dfrac{x}{{{{\left( {x + 1} \right)}^2}}} \Rightarrow \int {f'\left( x \right)dx} = \int {\dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx} \Rightarrow f\left( x \right) = \int {\left[ {\dfrac{{x + 1}}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]dx} \)
\( \Rightarrow f\left( x \right) = \ln \left| {x + 1} \right| + \dfrac{1}{{x + 1}} + C.\)
Ta lại có \(f\left( 1 \right) = \dfrac{1}{2} \Rightarrow \ln 2 + \dfrac{1}{2} + C = \dfrac{1}{2} \Rightarrow C = - \ln 2.\)
Trên khoảng \(\left( { - 1; + \infty } \right)\) ta có \(f\left( x \right) = \ln \left( {x + 1} \right) + \dfrac{1}{{x + 1}} - \ln 2.\)
Khi đó \(\int\limits_1^2 {f\left( x \right)dx} = \int\limits_1^2 {\ln \left( {x + 1} \right)dx} + \int\limits_1^2 {\dfrac{1}{{x + 1}}dx} + \int\limits_1^2 {\left( { - \ln 2} \right)dx} \)
\(\begin{array}{l} = \int\limits_1^2 {\ln \left( {x + 1} \right)d\left( {x + 1} \right)} + \left. {\left[ {\ln \left( {x + 1} \right) - x\ln 2} \right]} \right|_1^2\\ = \left. {\left( {x + 1} \right)\ln \left( {x + 1} \right)} \right|_1^2 - \int\limits_1^2 {1dx} + \left[ {\left( {\ln 3 - 2\ln 2} \right) - \left( {\ln 2 - \ln 2} \right)} \right]\\ = 4\ln 3 - 4\ln 2 - 1 - \ln 2 = 4\ln \dfrac{3}{2} - 1. \end{array}\)
Do đó \(a = 4,b = 3,c = 2,d = 1.\) Vậy \(a + b + c + d = 10\).