Question

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

• A. $\sqrt{e}+1$
• B. $\sqrt{e}-1$
• C. $\frac{\sqrt{e}+1}{e}$
• D. $\frac{\sqrt{e}-1}{e}$

Answer 1

The correct option is D $\frac{\sqrt{e}-1}{e}$

$Putt=-\frac{1}{x}$
$\Rightarrow dt=\frac{1}{x^2}dx$,then it reduces to
${\int }_{-1}^{-\frac{1}{2}}{e}^{t}dt=[e^t{]}_{-1}^{-\frac{1}{2}}=e^{-\frac{1}{2}}-e^{-1}=\frac{\sqrt{e}-1}{e}$