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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4xe^{3x}

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

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

Solution steps

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

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= 4 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Simplify

4 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : \frac{4}{9} (3 e^{3 x} x - e^{3 x})

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

Add a constant to the solution

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

e^{x} + e^{- x}

\frac{\partial}{\partial x} (cos (x + e^{y}))

\int \frac{9 x}{x - 1} d x

y = 3 x^{2} - 12 x + 1, (0, 1)

f (x) = x^{2} + 2

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