What is the integral of xe^{(2x)/3} ?

The integral of xe^{(2x)/3} is 9/4 (2/3 e^{(2x)/3}x-e^{(2x)/3})+C

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integral of xe^{(2x)/3}

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

\frac{9}{4} (\frac{2}{3} e^{\frac{2 x}{3}} x - e^{\frac{2 x}{3}}) + C

Solution steps

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

Apply u-substitution

= \int \frac{9 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{9}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{9}{4} (e^{u} u - \int e^{u} d u)

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

= \frac{9}{4} (e^{u} u - e^{u})

Substitute back

u = \frac{2 x}{3}

= \frac{9}{4} (e^{\frac{2 x}{3}} \frac{2 x}{3} - e^{\frac{2 x}{3}})

Simplify

\frac{9}{4} (e^{\frac{2 x}{3}} \frac{2 x}{3} - e^{\frac{2 x}{3}}) : \frac{9}{4} (\frac{2}{3} e^{\frac{2 x}{3}} x - e^{\frac{2 x}{3}})

= \frac{9}{4} (\frac{2}{3} e^{\frac{2 x}{3}} x - e^{\frac{2 x}{3}})

Add a constant to the solution

= \frac{9}{4} (\frac{2}{3} e^{\frac{2 x}{3}} x - e^{\frac{2 x}{3}}) + C

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What is the integral of xe^{(2x)/3} ?
The integral of xe^{(2x)/3} is 9/4 (2/3 e^{(2x)/3}x-e^{(2x)/3})+C