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

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

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

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

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

Solution steps

\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

\frac{d x}{d t} = x (100 - x)

\int e^{- \frac{x}{4}} d x

y = \frac{6 x}{x + 5}, (- 3, - 9)

x^{6} y^{^{'}} + x^{5} y = 0

\frac{d}{d x} (\frac{x}{x + 1})

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