Question

${\int }_1^2\left(\frac{x-1}{x^2}\right)e^{x}dx$

Answer 1

${\int }_1^2{e}^x\left(\frac{x-1}{x^2}\right)dx={\int }_1^2{e}^x(\frac{1}{x}-\frac{1}{x^2})dx$

If $f\left(x\right)=\frac{1}{x}$

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

$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)$

$\therefore {\int }_1^2{e}^x(\frac{1}{x}-\frac{1}{x^2})dx$

$={e^x\left(\frac{1}{x}\right)|}_1^2$

$=\frac{e^2}{2}-\frac{e^1}{1}=e^1(\frac{e}{2}-1)$

$=\frac{e(e-2)}{2}$.