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प्रश्न
उत्तर
\[Let\ I = \int_1^e \frac{e^x}{x}\left( 1 + x \log x \right)\ d\ x\ . Then, \]
\[I = \int_1^e \left( \frac{e^x}{x} + e^x \log x \right) dx\]
\[ \Rightarrow I = \int_1^e \frac{e^x}{x} dx + \int_1^e e^x \log x\ d\ x\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left[ \log x e^x \right]_1^e - \int_1^e e^x \log x d x + \int_1^e e^x \log\ x\ d\ x\]
\[ \Rightarrow I = \left( \log e \right) e^e - 0\]
\[ \Rightarrow I = e^e\]
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