integral of (2x+1)e^{0.4x}

解

\int (2 x + 1) e^{0.4 x} d x

5 e^{0.4 x} x - 10 e^{0.4 x} + C

解答ステップ

\int (2 x + 1) e^{0.4 x} d x

拡張

(2 x + 1) e^{0.4 x} : 2 e^{0.4 x} x + e^{0.4 x}

= \int 2 e^{0.4 x} x + e^{0.4 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 2 e^{0.4 x} x d x + \int e^{0.4 x} d x

\int 2 e^{0.4 x} x d x = 2 (2.5 e^{0.4 x} x - 6.25 e^{0.4 x})

\int e^{0.4 x} d x = 2.5 e^{0.4 x}

= 2 (2.5 e^{0.4 x} x - 6.25 e^{0.4 x}) + 2.5 e^{0.4 x}

簡素化

2 (2.5 e^{0.4 x} x - 6.25 e^{0.4 x}) + 2.5 e^{0.4 x} : 5 e^{0.4 x} x - 10 e^{0.4 x}

= 5 e^{0.4 x} x - 10 e^{0.4 x}

定数を解答に追加する

= 5 e^{0.4 x} x - 10 e^{0.4 x} + C