integral of (x^3-3x)e^{6x}

解

\int (x^{3} - 3 x) e^{6 x} d x

\frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x} + C

解答ステップ

\int (x^{3} - 3 x) e^{6 x} d x

拡張

(x^{3} - 3 x) e^{6 x} : e^{6 x} x^{3} - 3 e^{6 x} x

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

\int e^{6 x} x^{3} d x = \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)

\int 3 e^{6 x} x d x = \frac{1}{12} (6 e^{6 x} x - e^{6 x})

= \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) - \frac{1}{12} (6 e^{6 x} x - e^{6 x})

簡素化

\frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) - \frac{1}{12} (6 e^{6 x} x - e^{6 x}) : \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x}

= \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x}

定数を解答に追加する

= \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x} + C