Question
Integrate the function in Exercise:
$\text{e}^\text{x}\Bigg(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Bigg)$

Answer

$\text{I}=\int\text{e}^\text{x}\Bigg[\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Bigg]\text{dx}$
Also, let $\frac{1}{\text{x}}=\text{f}(\text{x})\Rightarrow \ \text{f}'(\text{x})=\frac{-1}{\text{x}^2}$
It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^\text{x}}{\text{x}}+\text{C}$

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