What is the integral of x^2e^{-18x} ?

The integral of x^2e^{-18x} is -1/2916 e^{-18x}(162x^2+18x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2e^{-18x}

\int x^{2} e^{- 18 x} d x

- \frac{1}{2916} e^{- 18 x} (162 x^{2} + 18 x + 1) + C

Solution steps

\int x^{2} e^{- 18 x} d x

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{5832} d u

= \int - \frac{e ^{u} u ^{2}}{5832} d u

Take the constant out:

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

= - \frac{1}{5832} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{5832} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{5832} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 18 x

= - \frac{1}{5832} ((- 18 x)^{2} e^{- 18 x} - 2 (e^{- 18 x} (- 18 x) - e^{- 18 x}))

Simplify

- \frac{1}{5832} ((- 18 x)^{2} e^{- 18 x} - 2 (e^{- 18 x} (- 18 x) - e^{- 18 x})) : - \frac{1}{2916} e^{- 18 x} (162 x^{2} + 18 x + 1)

= - \frac{1}{2916} e^{- 18 x} (162 x^{2} + 18 x + 1)

Add a constant to the solution

= - \frac{1}{2916} e^{- 18 x} (162 x^{2} + 18 x + 1) + C

(2 x^{4} + y^{4}) d y - x^{3} y d x = 0

f (x) = y

\frac{\partial}{\partial x} (\frac{m ^{2}}{4 x y})

\frac{d}{d x} (arcsin (1))

x \to 2 lim (2 x - \frac{1}{x})

What is the integral of x^2e^{-18x} ?
The integral of x^2e^{-18x} is -1/2916 e^{-18x}(162x^2+18x+1)+C