Đặt  u = ln (x+1)$\Rightarrow $ du = $\frac{1}{x+1}$dx

            dv = ${{x}^{2}}$dx $\Rightarrow $   v = $\frac{{{x}^{3}}}{3}$

$I=\int\limits_{1}^{e}{{{x}^{2}}.\ln (x+1)dx}$$\left. =\frac{{{x}^{3}}}{3}\ln (x+1) \right|_{0}^{1}$$-\frac{1}{3}\int\limits_{0}^{1}{{}}\frac{{{x}^{3}}}{x+1}dx$$=\frac{1}{3}\ln 2$$-\frac{1}{3}\int\limits_{0}^{1}{{}}\frac{{{x}^{3}}}{x+1}dx$

Tính $\int\limits_{0}^{1}{{}}\frac{{{x}^{3}}}{x+1}dx$$=\int\limits_{0}^{1}{\frac{{{x}^{3}}+1-1}{x+1}}dx$  $=\int\limits_{0}^{1}{\left( \left( {{x}^{2}}-x+1 \right)-\frac{1}{x+1} \right)}dx$$=\frac{5}{6}-\ln 2$

$I=\int\limits_{1}^{e}{{{x}^{2}}.\ln (x+1)dx}$$=\frac{1}{3}\ln 2$$-\frac{1}{3}\left( \frac{5}{6}-\ln 2 \right)$$=-\frac{5}{18}+\frac{2}{3}\ln 2$