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

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

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

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

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

Solution steps

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

Simplify

\frac{1}{x ^{2}} e^{- \frac{1}{x}} : \frac{e ^{- \frac{1}{x}}}{x ^{2}}

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

Apply u-substitution

: \int e^{u} d u

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