integral of (x-1)4^x

解

\int (x - 1) 4^{x} d x

\frac{2 ^{2 x - 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2 x - 2} - \frac{2 ^{2 x - 1}}{ln ( 2 )} + C

解答ステップ

\int (x - 1) \cdot 4^{x} d x

拡張

(x - 1) \cdot 4^{x} : 4^{x} x - 4^{x}

= \int 4^{x} x - 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 4^{x} x d x - \int 4^{x} d x

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

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

= \frac{2 ^{2 x - 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2 x - 2} - \frac{2 ^{2 x - 1}}{ln ( 2 )}

定数を解答に追加する

= \frac{2 ^{2 x - 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2 x - 2} - \frac{2 ^{2 x - 1}}{ln ( 2 )} + C