Lời giải:Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {2x + 1} \right)\\
dv = xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{2}{{2x + 1}}dx\\
v = \frac{{{x^2}}}{2}
\end{array} \right. \Rightarrow I = \left[ {\frac{{{x^2}}}{2}\ln \left( {2x + 1} \right)} \right]\left| {\mathop {}\limits_0^4 - \int\limits_0^4 {\frac{{{x^2}}}{{2x + 1}}dx} } \right.\)
\( \Leftrightarrow I = \left[ {\frac{{{x^2}}}{2}\ln \left( {2x + 1} \right)} \right]\left| {\mathop {}\limits_0^4 - \int\limits_0^4 {\left( {\frac{x}{2} - \frac{1}{4} + \frac{1}{{4\left( {2x + 1} \right)}}} \right)dx = } } \right.\left[ {\frac{{{x^2}}}{2}\ln \left( {2x + 1} \right)} \right]\left| {\mathop {}\limits_0^4 - \left( {\frac{{{x^2}}}{4} - \frac{1}{4}x + \frac{1}{8}\ln \left( {2x + 1} \right)} \right)\left| {\mathop {}\limits_0^4 } \right.} \right.\)
\( \Leftrightarrow I = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l}
a = 63\\
b = 4\\
c = 3
\end{array} \right. \Rightarrow S = a + b + c = 70\)
