integral of (4x^3+5x)e^{x^2}

解

\int (4 x^{3} + 5 x) e^{x^{2}} d x

2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C

解答ステップ

\int (4 x^{3} + 5 x) e^{x^{2}} d x

拡張

(4 x^{3} + 5 x) e^{x^{2}} : 4 e^{x^{2}} x^{3} + 5 e^{x^{2}} x

= \int 4 e^{x^{2}} x^{3} + 5 e^{x^{2}} 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 4 e^{x^{2}} x^{3} d x + \int 5 e^{x^{2}} x d x

\int 4 e^{x^{2}} x^{3} d x = 2 (e^{x^{2}} x^{2} - e^{x^{2}})

\int 5 e^{x^{2}} x d x = \frac{5}{2} e^{x^{2}}

= 2 (e^{x^{2}} x^{2} - e^{x^{2}}) + \frac{5}{2} e^{x^{2}}

簡素化

2 (e^{x^{2}} x^{2} - e^{x^{2}}) + \frac{5}{2} e^{x^{2}} : 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

= 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

定数を解答に追加する

= 2 e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C