What is the integral of 4xe^{(2x)/3} ?

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

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integral of 4xe^{(2x)/3}

\int 4 x e^{\frac{2 x}{3}} d x

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

Solution steps

\int 4 x e^{\frac{2 x}{3}} d x

Take the constant out:

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

= 4 \cdot \int x e^{\frac{2 x}{3}} d x

Apply u-substitution

: \int \frac{9 e ^{u} u}{4} d u

= 4 \cdot \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = \frac{2 x}{3}

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{\infty} e^{- 3 x} d x

\int_{3}^{\infty} \frac{1}{x - 2} d x

\int x^{2} + 2 x + 3 d x

\frac{d}{d x} ((x^{2} + y^{2})^{6})

\frac{\partial}{\partial v} (\frac{u + v}{2})

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