What is the integral of-2xe^{-4x} ?

The integral of-2xe^{-4x} is -1/8 (-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-2xe^{-4x}

\int - 2 x e^{- 4 x} d x

- \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

\int - 2 x e^{- 4 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^{- 4 x} d x

Apply u-substitution

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

= - 2 \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

= - 2 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= - 2 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= - 2 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 x

= - 2 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x})

Simplify

- 2 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x}) : - \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x})

= - \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x})

Add a constant to the solution

= - \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

\int \frac{1}{3 ( x - 1 ) ^{4}} d x

\int \frac{5}{t + 4} d t

\int 5 t^{5} d t

\int \frac{( 3 x ^{5} - 6 x ^{4} + 5 x ^{3} )}{x ^{4}} d x

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

What is the integral of-2xe^{-4x} ?
The integral of-2xe^{-4x} is -1/8 (-4e^{-4x}x-e^{-4x})+C