What is the integral of ye^{-2y} ?

The integral of ye^{-2y} is 1/4 (-2e^{-2y}y-e^{-2y})+C

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integral of ye^{-2y}

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

\frac{1}{4} (- 2 e^{- 2 y} y - e^{- 2 y}) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = - 2 y

= \frac{1}{4} (e^{- 2 y} (- 2 y) - e^{- 2 y})

Simplify

= \frac{1}{4} (- 2 e^{- 2 y} y - e^{- 2 y})

Add a constant to the solution

= \frac{1}{4} (- 2 e^{- 2 y} y - e^{- 2 y}) + C

\frac{d}{d t} (\frac{1}{2} t^{2} + t)

f (x) = 3^{- 6 x + 6}

\int_{0}^{2} \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\frac{d}{d x} (3 x^{2} - 6 x + 1)

\frac{1}{( s - 2 ) ^{2} - 4}

What is the integral of ye^{-2y} ?
The integral of ye^{-2y} is 1/4 (-2e^{-2y}y-e^{-2y})+C