Đáp án đúng là: D

Đặt \[\left\{ \begin{array}{l}u = \ln \left( {x + 3} \right)\\dv = \frac{{dx}}{{{x^2}}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{{x + 3}}\\v = \frac{1}{x}\end{array} \right.\]

Ta có: \[\int {\frac{{\ln \left( {x + 3} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 3} \right) + \int {\frac{{dx}}{{x\left( {x + 3} \right)}}} \]

Đặt \[I = \int {\frac{{dx}}{{x\left( {x + 3} \right)}}} = \frac{1}{3}\int {\left( {\frac{1}{x} - \frac{1}{{x + 3}}} \right)dx = \frac{1}{3}\ln \left| x \right| - \frac{1}{3}\ln \left| {x + 3} \right| = \frac{1}{3}\ln \left| {\frac{x}{{x + 3}}} \right|} \].

Suy ra \[F\left( x \right) = - \frac{1}{x}\ln \left( {x + 3} \right) + \frac{1}{3}\ln \left| {\frac{x}{{x + 3}}} \right| + C.\]

Lại có \[F\left( { - 2} \right) + F\left( 1 \right) = 0\]

\[\left( {\frac{1}{3}\ln 2 + C} \right) + \left( { - \ln 4 + \frac{1}{3}\ln \frac{1}{4} + C} \right) = 0\]

\[ \Leftrightarrow 2C = \frac{7}{3}\ln 2 \Leftrightarrow C = \frac{7}{6}\ln 2.\]

Suy ra \[F\left( { - 1} \right) + F\left( 2 \right) = \ln 2 + \frac{1}{3}\ln 2 - \frac{1}{2}\ln 5 + \frac{1}{3}\ln \frac{2}{5} + \frac{7}{3}\ln 2 = \frac{{10}}{3}\ln 2 - \frac{5}{6}\ln 5.\]