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

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

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integral of ye^{3y}

\int y e^{3 y} d y

\frac{1}{9} (e^{3 y} \cdot 3 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} \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

n = 0 \sum \infty \frac{1}{3 + 2 n}

\int \frac{sin ( x )}{3} d x

\int_{1}^{2} (x^{2} - 4) d x

\int 2 cos^{2} (t) sin^{2} (t) d t

f (x) = x^{2} + 3 x + 5, (1, 9)

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