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

Simplify

= \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}{d y} (- \frac{y}{( x + 2 ) ^{2}})

\frac{d}{d x} (ln (3 x) - e^{s e c (x)})

sin (tan (8 x))

f (x) = \frac{x + 1}{x ^{2} - 5}

f (t) = e^{t} cos (t)

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