What is the integral of e^{3y}y ?

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

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integral of e^{3y}y

\int e^{3 y} y d y

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

Solution steps

\int e^{3 y} 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} \cdot 3 y - e^{3 y})

Add a constant to the solution

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

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

\frac{d y}{d x} = 2 y x

\int \frac{2 ^{t}}{2 ^{t} + 3} d t

\frac{\partial}{\partial x} (7 e^{x^{8} y})

y^{^{'}} = 4 x - 6

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