Lời giải chi tiết:

\(\begin{array}{l}
I = \int\limits_0^1 {\frac{{{x^2} + 2x}}{{{{\left( {x + 1} \right)}^3}}}dx} = \int\limits_0^1 {\frac{{{x^2} + 2x + 1}}{{{{\left( {x + 1} \right)}^3}}}dx - \int\limits_0^1 {\frac{1}{{{{\left( {x + 1} \right)}^3}}}dx} } \\
= \int\limits_0^1 {\frac{1}{{x + 1}}dx + \left. {\frac{1}{{2{{\left( {x + 1} \right)}^2}}}} \right|} _0^1 = \left. {\ln \left| {x + 1} \right|} \right|_0^1 + \frac{1}{8} - \frac{1}{2}\\
= \ln 2 - \frac{3}{8}.\\
\Rightarrow \left\{ \begin{array}{l}
a = - \frac{3}{8}\\
b = 1
\end{array} \right. \Rightarrow 16a + b = - 5.
\end{array}\)

Chọn D.