Question

\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]

Answer 1

\[\int_1^2 \frac{1}{x^2} e^\frac{- 1}{x} d x\]

\[Let \frac{- 1}{x} = t, then \frac{1}{x^2} dx = dt\]

\[\text{When, }x \to 1 ; t \to - 1\]

\[\text{And }x \to 2 ; t \to \frac{- 1}{2}\]

Therefore the integral becomes

\[ \int_{- 1}^\frac{- 1}{2} e^t d t\]

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

\[ = e^\frac{- 1}{2} - e^{- 1} \]

\[ = \frac{\sqrt{e} - 1}{e}\]