Question:
If $\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x=f(x) e^{x}+C$, then write the value of $f(x)$.
Solution:
Consider, $\int \frac{x-1}{x^{2}} e^{x} \mathrm{dx}$
$=\int \frac{x}{x^{2}}-\frac{1}{x^{2}} e^{x} d x$
$=\int \frac{1}{x}-\frac{1}{x^{2}} e^{x} d x$
It is clearly of the form,
$\int e^{x}\left[f(x)+f^{l}(x)\right] d x=e^{x} f(x)+c$
By comparison, $\mathrm{f}(\mathrm{x})=\frac{1}{x} ; \mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{x^{2}}$
$=e^{x} \frac{1}{x}+c$
Therefore, the value of $f(x)=\frac{1}{x}$
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