Question

\[\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx\]

Answer 1

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