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प्रश्न
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उत्तर
\[\int\left( \frac{x - 1}{x^2} \right) e^x dx = \int\left( \frac{x}{x^2} - \frac{1}{x^2} \right) e^x dx\]
\[ = \int\left( \frac{1}{x} - \frac{1}{x^2} \right) e^x dx\]
\[\text{ Consider,} f\left( x \right) = \frac{1}{x},\text{ then f}^ \left( x \right) = - \frac{1}{x^2}\]
\[\text{ Thus , the given integrand is of the form e}^x \left[ f\left( x \right) + f^ \left( x \right) \right] . \]
\[\text{ Therefore, }\int\left( \frac{x - 1}{x^2} \right) e^x dx = \frac{1}{x} e^x + C\]
\[\text{ Hence,} f\left( x \right) = \frac{1}{x} .\]
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