integral of (x^2+x)/(e^x)

解

\int \frac{x ^{2} + x}{e ^{x}} d x

- \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}} + C

解答ステップ

\int \frac{x ^{2} + x}{e ^{x}} d x

拡張

\frac{x ^{2} + x}{e ^{x}} : \frac{x ^{2}}{e ^{x}} + \frac{x}{e ^{x}}

= \int \frac{x ^{2}}{e ^{x}} + \frac{x}{e ^{x}} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{e ^{x}} d x + \int \frac{x}{e ^{x}} d x

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

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

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

簡素化

- \frac{x ^{2}}{e ^{x}} + 2 (- \frac{x}{e ^{x}} - \frac{1}{e ^{x}}) - \frac{x}{e ^{x}} - \frac{1}{e ^{x}} : - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}}

= - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}}

定数を解答に追加する

= - \frac{x ^{2}}{e ^{x}} - \frac{2 x}{e ^{x}} - \frac{x}{e ^{x}} - \frac{3}{e ^{x}} + C