What is the integral from 0 to 1 of (e^{-1/x})/(x^2) ?

The integral from 0 to 1 of (e^{-1/x})/(x^2) is 1/e

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integral from 0 to 1 of (e^{-1/x})/(x^2)

\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

\frac{1}{e}

Solution steps

\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply Integration By Parts

: [\frac{e ^{- \frac{1}{x}}}{x} - \int \frac{e ^{- \frac{1}{x}} - 2 e ^{- \frac{1}{x}} x}{x ^{3}} d x]_{0}^{1}

= [\frac{e ^{- \frac{1}{x}}}{x} - \int \frac{e ^{- \frac{1}{x}} - 2 e ^{- \frac{1}{x}} x}{x ^{3}} d x]_{0}^{1}

\int \frac{e ^{- \frac{1}{x}} - 2 e ^{- \frac{1}{x}} x}{x ^{3}} d x = \frac{e ^{- \frac{1}{x}}}{x} - e^{- \frac{1}{x}}

= [\frac{e ^{- \frac{1}{x}}}{x} - (\frac{e ^{- \frac{1}{x}}}{x} - e^{- \frac{1}{x}})]_{0}^{1}

Simplify

= [e^{- \frac{1}{x}}]_{0}^{1}

Compute the boundaries:

\frac{1}{e}

= \frac{1}{e}

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What is the integral from 0 to 1 of (e^{-1/x})/(x^2) ?
The integral from 0 to 1 of (e^{-1/x})/(x^2) is 1/e