integral of (3-x^2)2^x

解

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

\frac{3 \cdot 2 ^{x}}{ln ( 2 )} - \frac{2 ^{x} x ^{2}}{ln ( 2 )} + \frac{1}{ln ^{2} ( 2 )} \cdot 2^{x + 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{1 + x} + C

解答ステップ

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

拡張

(3 - x^{2}) \cdot 2^{x} : 3 \cdot 2^{x} - 2^{x} x^{2}

= \int 3 \cdot 2^{x} - 2^{x} x^{2} 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 3 \cdot 2^{x} d x - \int 2^{x} x^{2} d x

\int 3 \cdot 2^{x} d x = \frac{3 \cdot 2 ^{x}}{ln ( 2 )}

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

= \frac{3 \cdot 2 ^{x}}{ln ( 2 )} - (\frac{2 ^{x} x ^{2}}{ln ( 2 )} - \frac{1}{ln ( 2 )} (\frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x}))

簡素化

\frac{3 \cdot 2 ^{x}}{ln ( 2 )} - (\frac{2 ^{x} x ^{2}}{ln ( 2 )} - \frac{1}{ln ( 2 )} (\frac{2 ^{x + 1} x}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{1 + x})) : \frac{3 \cdot 2 ^{x}}{ln ( 2 )} - \frac{2 ^{x} x ^{2}}{ln ( 2 )} + \frac{1}{ln ^{2} ( 2 )} \cdot 2^{x + 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{1 + x}

= \frac{3 \cdot 2 ^{x}}{ln ( 2 )} - \frac{2 ^{x} x ^{2}}{ln ( 2 )} + \frac{1}{ln ^{2} ( 2 )} \cdot 2^{x + 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{1 + x}

定数を解答に追加する

= \frac{3 \cdot 2 ^{x}}{ln ( 2 )} - \frac{2 ^{x} x ^{2}}{ln ( 2 )} + \frac{1}{ln ^{2} ( 2 )} \cdot 2^{x + 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{1 + x} + C