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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - y e^{u} d u

= \frac{1}{y} \cdot \int - y e^{u} d u

Take the constant out:

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

= \frac{1}{y} (- y \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{y} (- y e^{u})

Substitute back

u = - \frac{x}{y}

= \frac{1}{y} (- y e^{- \frac{x}{y}})

Simplify

\frac{1}{y} (- y e^{- \frac{x}{y}}) : - e^{- \frac{x}{y}}

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

Add a constant to the solution

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

sin (x^{2})

\frac{\partial}{\partial x} (y^{3} sin (x + y))

\frac{d}{d x} (\frac{x}{x ^{- 1} + 9})

\int (\frac{3 x ^{2}}{64 - x ^{2}}) d x

x \to 0 + lim (\frac{x ^{- 2}}{x})

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