What is the integral of 2/3 ye^y ?

The integral of 2/3 ye^y is 2/3 (e^yy-e^y)+C

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integral of 2/3 ye^y

\int \frac{2}{3} y e^{y} d y

\frac{2}{3} (e^{y} y - e^{y}) + C

Solution steps

\int \frac{2}{3} y e^{y} d y

Take the constant out:

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

= \frac{2}{3} \cdot \int y e^{y} d y

Apply Integration By Parts

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

= \frac{2}{3} (e^{y} y - \int e^{y} d y)

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

= \frac{2}{3} (e^{y} y - e^{y})

Add a constant to the solution

= \frac{2}{3} (e^{y} y - e^{y}) + C

\frac{d}{d x} (- 2 x^{3} - 7 x + 5)

\frac{\partial}{\partial x} (y^{0.25} x^{0.75})

f (x) = x^{π}

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

\int (\frac{t ^{3}}{3} + \frac{1}{4 t ^{2}}) d t