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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{9 e ^{u} u}{16} d u

= 4 \cdot \int \frac{9 e ^{u} u}{16} d u

Take the constant out:

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

= 4 \cdot \frac{9}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{9}{16} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{9}{16} (e^{u} u - e^{u})

Substitute back

u = \frac{4 x}{3}

= 4 \cdot \frac{9}{16} (e^{\frac{4 x}{3}} \frac{4 x}{3} - e^{\frac{4 x}{3}})

Simplify

4 \cdot \frac{9}{16} (e^{\frac{4 x}{3}} \frac{4 x}{3} - e^{\frac{4 x}{3}}) : 3 e^{\frac{4 x}{3}} x - \frac{9}{4} e^{\frac{4 x}{3}}

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

Add a constant to the solution

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

\int x (2 x - 1) d x

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

x \to 1 lim (x (2))

\frac{d}{d x} (a x^{2} + b y (x)^{2} + λ (x^{2} + y^{2} - c^{2}))

x^{2} - 2 x y + x y^{^{'}} = 0

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