integral of (x^2)/a e^{-x/a}

解答

\int \frac{x ^{2}}{a} e^{- \frac{x}{a}} d x

- e^{- \frac{x}{a}} (x^{2} + 2 a (x + a)) + C

求解步骤

\int \frac{x ^{2}}{a} e^{- \frac{x}{a}} d x

提出常数:

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

= \frac{1}{a} \cdot \int x^{2} e^{- \frac{x}{a}} d x

使用分布积分法

= \frac{1}{a} (- a e^{- \frac{x}{a}} x^{2} - \int - 2 a e^{- \frac{x}{a}} x d x)

\int - 2 a e^{- \frac{x}{a}} x d x = - 2 a (- a e^{- \frac{x}{a}} x - a^{2} e^{- \frac{x}{a}})

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

化简

\frac{1}{a} (- a e^{- \frac{x}{a}} x^{2} - (- 2 a (- a e^{- \frac{x}{a}} x - a^{2} e^{- \frac{x}{a}}))) : - e^{- \frac{x}{a}} (x^{2} + 2 a (x + a))

= - e^{- \frac{x}{a}} (x^{2} + 2 a (x + a))

解答补常数

= - e^{- \frac{x}{a}} (x^{2} + 2 a (x + a)) + C

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