Question

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

Answer 1

$\begin{array}{c}{\int }_1^2{e}^x\frac{\left(x-1\right)}{x^2}dx={\int }_1^2{e}^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx\\ ={\int }_1^2{e}^x\frac{1}{x}dx-{\int }_1^2{e}^x\frac{1}{x^2}dx\\ =\frac{1}{x}{\int }_1^2{e}^{x}dx-{\int }_1^2\left(\frac{d}{dx}\cdot \frac{1}{x}\cdot {\int }_1^2{e}^{x}dx\right)dx-{\int }_1^2{e}^x\frac{1}{x^2}dx\\ =\frac{1}{x{e}^x}-{\int }_1^2\frac{1}{x^2}{e}^{x}dx-{\int }_1^2{e}^x\frac{1}{x^2}dx\\ =\frac{e^x}{x}+{\int }_1^2{e}^x\frac{1}{x^2}dx-{\int }_1^2{e}^x\frac{1}{x^2}dx+c\\ =\frac{e^2}{2}-e+C\end{array}$

Hence, this is the answer.