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

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

\int (2 x e^{- 3 x}) d x

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 x

= 2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x})

Simplify

2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x}) : \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x})

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

Add a constant to the solution

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

x \to 0 lim (\frac{3 tan ( x )}{x})

y^{^{'}} + 2 sin (2 x) y = 2 sin (2 x), y (\frac{π}{2}) = 3

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

\frac{d}{d x} (- csc (x) - sin (x))

\int 5 sin^{3} (9 x) d x

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