What is the integral of ye^{-3y} ?

The integral of ye^{-3y} is 1/9 (-3e^{-3y}y-e^{-3y})+C

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integral of ye^{-3y}

\int y e^{- 3 y} d y

\frac{1}{9} (- 3 e^{- 3 y} y - e^{- 3 y}) + C

Solution steps

\int y e^{- 3 y} d y

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 y

= \frac{1}{9} (e^{- 3 y} (- 3 y) - e^{- 3 y})

Simplify

= \frac{1}{9} (- 3 e^{- 3 y} y - e^{- 3 y})

Add a constant to the solution

= \frac{1}{9} (- 3 e^{- 3 y} y - e^{- 3 y}) + C

\frac{d}{d x} ((x^{2} + 3 x)^{4})

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\frac{1}{z}

What is the integral of ye^{-3y} ?
The integral of ye^{-3y} is 1/9 (-3e^{-3y}y-e^{-3y})+C