Question

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

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

Answer 1

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

Now,

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

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

$={\left(\frac{e^x}{x}\right)|}_1^2$$+\underset{1}{\overset{2}{\int }}{e}^x\left(\frac{1}{x^2}\right)dx$$-\underset{1}{\overset{2}{\int }}{e}^x\left(\frac{1}{x^2}\right)dx$ [ Using method of by parts]

$=\frac{e^2}{2}-e$

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