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

The integral of 3xe^{-4x} is 3/16 (-4e^{-4x}x-e^{-4x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3xe^{-4x}

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

\frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 x

= 3 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x})

Simplify

3 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x}) : \frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x})

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

Add a constant to the solution

= \frac{3}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

\int \frac{8 x ^{2} + 5 x + 77}{( x + 1 ) ( x ^{2} + 9 )} d x

\int \frac{7 x + 5}{( x - 2 ) ( 1 - x )} d x

\int \frac{1}{cos ^{3} ( x ) sin ( x )} d x

\int \frac{( e ^{2 x} + 1 ) ^{2}}{e ^{x}} d x

\int \frac{( 1 + x ) ^{2}}{x + x ^{3}} d x

What is the integral of 3xe^{-4x} ?
The integral of 3xe^{-4x} is 3/16 (-4e^{-4x}x-e^{-4x})+C