What is the integral of θ/(y^2)e^{-θ/y} ?

The integral of θ/(y^2)e^{-θ/y} is -e^{θ/y}+C

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integral of θ/(y^2)e^{-θ/y}

\int \frac{θ}{y ^{2}} e^{- \frac{θ}{y}} d y

- e^{\frac{θ}{y}} + C

Solution steps

\int \frac{θ}{y ^{2}} e^{- \frac{θ}{y}} d y

Take the constant out:

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

= θ \cdot \int \frac{1}{y ^{2}} e^{- \frac{θ}{y}} d y

- \frac{θ}{y} = θ \frac{1}{y}

= θ \cdot \int \frac{1}{y ^{2}} e^{θ \frac{1}{y}} d y

Simplify

\frac{1}{y ^{2}} e^{θ \frac{1}{y}} : \frac{e ^{\frac{θ}{y}}}{y ^{2}}

= θ \cdot \int \frac{e ^{\frac{θ}{y}}}{y ^{2}} d y

Apply u-substitution

: \int - \frac{e ^{u}}{θ} d u

= θ \cdot \int - \frac{e ^{u}}{θ} d u

Take the constant out:

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

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

Use the common integral:

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

= θ (- \frac{1}{θ} e^{u})

Substitute back

u = \frac{θ}{y}

= θ (- \frac{1}{θ} e^{\frac{θ}{y}})

Simplify

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

= - e^{\frac{θ}{y}}

Add a constant to the solution

= - e^{\frac{θ}{y}} + C

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