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

解

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

\frac{2 ^{2 x - 1}}{ln ( 2 )} - 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{x ^{5}}{5} + C

解答ステップ

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

拡張

(2^{x} - x^{2})^{2} : 4^{x} - 2^{1 + x} x^{2} + x^{4}

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

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

\int 2^{1 + x} x^{2} d x = 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}))

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{2 ^{2 x - 1}}{ln ( 2 )} - 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{x ^{5}}{5}

定数を解答に追加する

= \frac{2 ^{2 x - 1}}{ln ( 2 )} - 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{x ^{5}}{5} + C