Biết rằng $\int\limits_{0}^{6}{\dfrac{{{x}^{3}}}{x+1}dx=a+b\ln...

Ta có $\int\limits_{0}^{6}{\dfrac{{{x}^{3}}}{x+1}dx}=\int\limits_{0}^{6}{\dfrac{{{x}^{2}}+1-1}{x+1}dx}=\int\limits_{0}^{6}{\left( {{x}^{2}}-x+1-\dfrac{1}{x+1} \right)dx}$
$=\left( \dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+x-\ln \left| x+1 \right| \right)\left| _{0}^{6} \right.=60-\ln 7\Rightarrow \left\{ \begin{aligned}
& a=60 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow S=58.$