integral of (5x^3+6x)e^{x^2}

解

\int (5 x^{3} + 6 x) e^{x^{2}} d x

\frac{5}{2} e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C

解答ステップ

\int (5 x^{3} + 6 x) e^{x^{2}} d x

拡張

(5 x^{3} + 6 x) e^{x^{2}} : 5 e^{x^{2}} x^{3} + 6 e^{x^{2}} x

= \int 5 e^{x^{2}} x^{3} + 6 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 5 e^{x^{2}} x^{3} d x + \int 6 e^{x^{2}} x d x

\int 5 e^{x^{2}} x^{3} d x = \frac{5}{2} (e^{x^{2}} x^{2} - e^{x^{2}})

\int 6 e^{x^{2}} x d x = 3 e^{x^{2}}

= \frac{5}{2} (e^{x^{2}} x^{2} - e^{x^{2}}) + 3 e^{x^{2}}

簡素化

\frac{5}{2} (e^{x^{2}} x^{2} - e^{x^{2}}) + 3 e^{x^{2}} : \frac{5}{2} e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

= \frac{5}{2} e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}}

定数を解答に追加する

= \frac{5}{2} e^{x^{2}} x^{2} + \frac{1}{2} e^{x^{2}} + C