What is the integral of (2p^3)/(e^p) ?

The integral of (2p^3)/(e^p) is 2(-e^{-p}p^3-3(e^{-p}p^2-2(-e^{-p}p-e^{-p})))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2p^3)/(e^p)

\int \frac{2 p ^{3}}{e ^{p}} d p

2 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p}))) + C

Solution steps

\int \frac{2 p ^{3}}{e ^{p}} d p

Take the constant out:

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

= 2 \cdot \int \frac{p ^{3}}{e ^{p}} d p

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{p ^{3}}{e ^{p}} = (p^{3}) e^{- p}

= 2 \cdot \int p^{3} e^{- p} d p

Apply u-substitution

: \int e^{u} u^{3} d u

= 2 \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= 2 (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= 2 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - p

= 2 ((- p)^{3} e^{- p} - 3 ((- p)^{2} e^{- p} - 2 (e^{- p} (- p) - e^{- p})))

Simplify

2 ((- p)^{3} e^{- p} - 3 ((- p)^{2} e^{- p} - 2 (e^{- p} (- p) - e^{- p}))) : 2 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p})))

= 2 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p})))

Add a constant to the solution

= 2 (- e^{- p} p^{3} - 3 (e^{- p} p^{2} - 2 (- e^{- p} p - e^{- p}))) + C

\int \frac{a - y}{y} d y

x^{3} + x

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

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

\frac{\partial}{\partial x} (x y - z^{2})

What is the integral of (2p^3)/(e^p) ?
The integral of (2p^3)/(e^p) is 2(-e^{-p}p^3-3(e^{-p}p^2-2(-e^{-p}p-e^{-p})))+C