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प्रश्न
उत्तर
\[Let\ I \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2 x^2} \right) d\ x\ . Then, \]
\[I = \int_1^2 e^{2x} \frac{1}{x} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{e^{2x}}{2x} \right]_1^2 - \int_1^2 - e^{2x} \frac{1}{2 x^2} \right\} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^{2x}}{2x} \right]_1^2 \]
\[ \Rightarrow I = \frac{e^4}{4} - \frac{e^2}{2}\]
\[ \Rightarrow I = \frac{e^4 - 2 e^2}{4}\]
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