Question

$\underset{1}{\overset{e}{\int }}\frac{e^x}{x}\left(1+x\log x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_1^e\frac{e^x}{x}\left(1+x\log x\right)\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_1^e\left(\frac{e^x}{x}+e^x\log x\right)dx \\ \Rightarrow I={\int }_1^e\frac{e^x}{x}dx+{\int }_1^e{e}^x\log x\mathit{d}x \\ \mathrm{Integrating}\mathrm{first}\mathrm{term}\mathrm{by}\mathrm{parts} \\ \Rightarrow I={\left[\log x{e}^x\right]}_1^e-{\int }_1^e{e}^x\log x\mathit{d}x+{\int }_1^e{e}^x\log x\mathit{d}x \\ \Rightarrow I=\left(\log e\right)e^e-0 \\ \Rightarrow I=e^e \end{gathered}$$