Question

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

Answer 1

$\int \frac{x-1}{x^2}\times e^{x}dx=\int e^x(\frac{1}{x}-\frac{1}{x^2})dx$

This is in the form of $\int e^x\left(g\right(x)+g^{\prime }(x\left)\right)dx=e^{x}g\left(x\right)+C$

Here $g\left(x\right)=\frac{1}{x}$ and $g^{\prime }\left(x\right)=-\frac{1}{x^2}$

$⟹\int \left(\frac{x-1}{x^2}\right)e^{x}dx=\frac{e^x}{x}+C$

$⟹f\left(x\right)=\frac{1}{x}$