Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{1}{\overset{e}{\int }}\frac{e^x}{x}\left(1+x\log x\right)dx \\ =\underset{1}{\overset{e}{\int }}{e}^x\left(\frac{1}{x}+\log x\right)dx \\ =\underset{1}{\overset{e}{\int }}{e}^x\left(\log x+\frac{1}{x}\right)dx \\ \mathrm{We}\mathrm{know}, \\ \int e^x\left(f\left(x\right)+f\text{'}\left(x\right)\right)dx=e^{x}f\left(x\right)+c \\ \mathrm{Here}, \\ f\left(x\right)=\log x \\ \Rightarrow f\text{'}\left(x\right)=\frac{1}{x} \\ \mathrm{Thus}, \\ I={\left[e^x\log x\right]}_1^e \\ =\left[e^e\left(\log e\right)-e^1\left(\log 1\right)\right] \\ =e^e\left(1\right)-e\left(0\right) \\ =e^e \end{gathered}$$

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Hence, $\underset{1}{\overset{e}{\int }}\frac{e^x}{x}\left(1+x\log x\right)dx=\underline{e^e}$.