b)
Đặt : $\left\{ \begin{array}{l} x=\ln x \Rightarrow du=\frac{dx}{x} \\ dv=(1+x)dx \Rightarrow v=x+\frac{x^2}{2} \end{array} \right. $
Lúc đó : $I=(x+\frac{x^2}{2} )\ln x \left|\begin{array}{l}e\\1\end{array}\right. -\int\limits_{1}^{e}(x^2+\frac{x^2}{2} )(\frac{1}{x} )dx$
$=(e^2+\frac{e^2}{2} )\ln e-0-\int\limits_{1}^{e}dx-\int\limits_{1}^{e}\frac{xdx}{2} $
$=(e+\frac{e^2}{2} )-x \left|\begin{array}{l}e\\1\end{array}\right. -\frac{x^2}{4} \left|\begin{array}{l}e\\1\end{array}\right. $
$=e+\frac{e^2}{2}-e(e-1)-\frac{1}{4}(e^2-1)=\frac{1}{4}(e^2+5)$