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

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

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

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

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

Solution steps

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

3 x - 8

(5 y - 2 x) y^{^{'}} - 2 y = 0

f (x) = x x^{2} + 60, at x = 2

\frac{\partial}{\partial x} ((3 - x) (3 - y) (x + y - 3))

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 3 x})

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