What is the integral from-7 to 0 of (e^{1/x})/(x^2) ?

The integral from-7 to 0 of (e^{1/x})/(x^2) is 1/(e^{1/7)}

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integral from-7 to 0 of (e^{1/x})/(x^2)

\int_{- 7}^{0} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

\frac{1}{e ^{\frac{1}{7}}}

Decimal Notation

0.86687 \dots

Solution steps

\int_{- 7}^{0} \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]_{- 7}^{0}

= [\frac{e ^{\frac{1}{x}}}{x} - \int \frac{- e ^{\frac{1}{x}} - 2 e ^{\frac{1}{x}} x}{x ^{3}} d x]_{- 7}^{0}

\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}})]_{- 7}^{0}

Simplify

= [- e^{\frac{1}{x}}]_{- 7}^{0}

Compute the boundaries:

\frac{1}{e ^{\frac{1}{7}}}

= \frac{1}{e ^{\frac{1}{7}}}

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What is the integral from-7 to 0 of (e^{1/x})/(x^2) ?
The integral from-7 to 0 of (e^{1/x})/(x^2) is 1/(e^{1/7)}