Question

${\int }_1^2{e}^x(\frac{1}{x}-\frac{1}{x^2})dx=$ [MNR 1990; AMU 1999; UPSEAT 2000; Pb. CET 2004]

• A. $\frac{e^2}{2}+e$
• B. $e-\frac{e^2}{2}$
• C. $\frac{e^2}{2}-e$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{e^2}{2}-e$

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