integral of x^2e^{-12x}

解答

\int x^{2} e^{- 12 x} d x

- \frac{1}{864} e^{- 12 x} (72 x^{2} + 12 x + 1) + C

求解步骤

\int x^{2} e^{- 12 x} d x

使用换元积分法

= \int - \frac{e ^{u} u ^{2}}{1728} d u

提出常数:

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

= - \frac{1}{1728} \cdot \int e^{u} u^{2} d u

使用分布积分法

= - \frac{1}{1728} (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}{1728} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

u = - 12 x

代回

= - \frac{1}{1728} ((- 12 x)^{2} e^{- 12 x} - 2 (e^{- 12 x} (- 12 x) - e^{- 12 x}))

化简

- \frac{1}{1728} ((- 12 x)^{2} e^{- 12 x} - 2 (e^{- 12 x} (- 12 x) - e^{- 12 x})) : - \frac{1}{864} e^{- 12 x} (72 x^{2} + 12 x + 1)

= - \frac{1}{864} e^{- 12 x} (72 x^{2} + 12 x + 1)

解答补常数

= - \frac{1}{864} e^{- 12 x} (72 x^{2} + 12 x + 1) + C

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