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

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

\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

使用分布积分法

= - \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}))

u = - 18 x

代回

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

化简

- \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)

解答补常数

= - \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

d er i v a t i v e 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})