Question

\[\text{ If } \int\left( \frac{x - 1}{x^2} \right) e^x dx = f\left( x \right) e^x + C, \text{ then  write  the value of  f}\left( x \right) .\]

Answer 1

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