What is the integral of e^{-u}u ?

The integral of e^{-u}u is -e^{-u}u-e^{-u}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-u}u

\int e^{- u} u d u

- e^{- u} u - e^{- u} + C

Solution steps

\int e^{- u} u d u

Apply Integration By Parts

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

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

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

= - e^{- u} u - e^{- u}

Add a constant to the solution

= - e^{- u} u - e^{- u} + C

x sin (\frac{8}{x})

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

\int \frac{2 x + 5}{( x - 4 ) ( x + 1 )} d x

y = 3 x^{3}

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (t)