Question

$\mathrm{If}\int \left(\frac{x-1}{x^2}\right){\mathrm{e}}^x\mathrm{d}x=f\left(x\right){\mathrm{e}}^x+C,\mathrm{then}\mathrm{write}\mathrm{the}\mathrm{value}\mathrm{of}f\left(x\right).$

Answer 1

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