What is the integral of x(1/4 e^{-x/4}) ?

The integral of x(1/4 e^{-x/4}) is -e^{-x/4}(x+4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(1/4 e^{-x/4})

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

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

Solution steps

\int x \frac{1}{4} e^{- \frac{x}{4}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int x e^{- \frac{x}{4}} d x

Apply Integration By Parts

: - 4 e^{- \frac{x}{4}} x - \int - 4 e^{- \frac{x}{4}} d x

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

\int - 4 e^{- \frac{x}{4}} d x = 16 e^{- \frac{x}{4}}

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

Simplify

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

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

Add a constant to the solution

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

2 x^{2}, x^{2} + 5

y^{^{''}} + 6 y^{^{'}} + 13 = 0

\int \frac{1}{16} e^{4 x} d x

\frac{d}{d x} (\frac{x ^{2}}{sin ( a x )})

\frac{x ^{2}}{( 5 + x ) ^{4}}

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