Question

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

Answer 1

${\int }_1^e(\frac{e^x}{x}+\frac{e^x}{x}x\log x)dx$

$={\int }_1^e(\frac{e^x}{x}+e^x\log x)dx$

$={\int }_1^e\frac{e^x}{x}dx+{\int }_1^e{e}^x\log xdx$

Consider ${\int }_1^e{e}^x\log xdx$

Let $u=\log x\Rightarrow du=\frac{1}{x}dx$

$dv=e^{x}dx\Rightarrow v=e^x$

$={\int }_1^e\frac{e^x}{x}dx+{\left[e^x\log x\right]}_1^e-{\int }_1^e\frac{e^x}{x}dx$

$={\left[e^x\log x\right]}_1^e$

$=[e^e\log e-e^1\log 1]$

$=e^e\log e$