Question

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

• 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 $\frac{\sqrt{e}-1}{e}$

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

putting $\frac{-1}{x}=t$

$\frac{1}{{}^{x^2}}dx=dt$

$={\int }_{-1}^{\frac{-1}{2}}{e}^{t}dt$

$={\left[e^t\right]}_{-1}^{\frac{-1}{2}}$

$=e^{\frac{-1}{2}}-e^{-1}$

$=\frac{1}{\sqrt{e}}-\frac{1}{e}$

$=\frac{e-\sqrt{e}}{e\sqrt{e}}$

$=\frac{\sqrt{e}-1}{e}$