integral of (x^2-6x)3^{4x}

解

\int (x^{2} - 6 x) 3^{4 x} d x

\frac{8 1 ^{x} x ^{2}}{4 ln ( 3 )} + \frac{- 1 - 12 ln ( 3 )}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )}) + C

解答ステップ

\int (x^{2} - 6 x) \cdot 3^{4 x} d x

拡張

(x^{2} - 6 x) \cdot 3^{4 x} : 8 1^{x} x^{2} - 2 \cdot 3^{1 + 4 x} x

= \int 8 1^{x} x^{2} - 2 \cdot 3^{1 + 4 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 8 1^{x} x^{2} d x - \int 2 \cdot 3^{1 + 4 x} x d x

\int 8 1^{x} x^{2} d x = \frac{8 1 ^{x} x ^{2}}{4 ln ( 3 )} - \frac{1}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )})

\int 2 \cdot 3^{1 + 4 x} x d x = 6 (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )})

= \frac{8 1 ^{x} x ^{2}}{4 ln ( 3 )} - \frac{1}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )}) - 6 (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )})

類似した元を足す:

- \frac{1}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )}) - 6 (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )}) = \frac{- 12 ln ( 3 ) - 1}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )})

= \frac{8 1 ^{x} x ^{2}}{4 ln ( 3 )} + \frac{- 1 - 12 ln ( 3 )}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )})

定数を解答に追加する

= \frac{8 1 ^{x} x ^{2}}{4 ln ( 3 )} + \frac{- 1 - 12 ln ( 3 )}{2 ln ( 3 )} (\frac{8 1 ^{x} x}{4 ln ( 3 )} - \frac{8 1 ^{x}}{16 ln ^{2} ( 3 )}) + C