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

The integral of (e^{x^{-1}})/(x^2) is -e^{1/x}+C

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

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

- e^{\frac{1}{x}} + C

Solution steps

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

x^{- 1} = \frac{1}{x}

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

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - e^{u}

Substitute back

u = \frac{1}{x}

= - e^{\frac{1}{x}}

Add a constant to the solution

= - e^{\frac{1}{x}} + C

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What is the integral of (e^{x^{-1}})/(x^2) ?
The integral of (e^{x^{-1}})/(x^2) is -e^{1/x}+C