What is the integral from 0 to 9 of (e^{1/x})/(x^2) ?

The integral from 0 to 9 of (e^{1/x})/(x^2) is diverges

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integral from 0 to 9 of (e^{1/x})/(x^2)

\int_{0}^{9} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{9} \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}^{9}

\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}^{9}

Simplify

= [- e^{\frac{1}{x}}]_{0}^{9}

Compute the boundaries:

diverges

= diverges

What is the integral from 0 to 9 of (e^{1/x})/(x^2) ?
The integral from 0 to 9 of (e^{1/x})/(x^2) is diverges