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

The integral of (xe^{-3x}) is 1/9 (-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 (xe^{-3x})

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

\frac{1}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - 3 x

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

Simplify

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

Add a constant to the solution

= \frac{1}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

\int cot (29 x) d x

\int 8 csc^{2} (4 x) cot (4 x) d x

\int e^{(1 + i) t} d t

\int x sec^{2} (x^{2}) tan (x^{2}) d x

\int \frac{x ^{3} - 2 x ^{2} + x + 1}{x ^{2} + 4 x + 3} d x

What is the integral of (xe^{-3x}) ?
The integral of (xe^{-3x}) is 1/9 (-3e^{-3x}x-e^{-3x})+C