Question

$\stackrel{2}{\underset{1}{\int }}{e}^x(\frac{1}{x}-\frac{1}{x^2})dx$ equals

• A. $e(\frac{e}{2}-1)$
• B. $1$
• C. $e(e-1)$
• D. $\frac{e}{2}$

Answer 1

The correct option is A $e(\frac{e}{2}-1)$

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

$\int \frac{e^x}{x}dx-\int e^x\frac{1}{x^2}dx$

Using Integration by parts we get

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

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

$=\frac{e^x}{x}+\int \frac{e^x}{x^2}dx-\int \frac{e^x}{x^2}dx$

$=\frac{e^x}{x}$

Evaluating the above integral over limits 1 to 2 we get

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

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