integral of 2(2x+1)^9

解

\int 2 (2 x + 1)^{9} d x

2 (\frac{256 x ^{10}}{5} + 256 x^{9} + 576 x^{8} + 768 x^{7} + 672 x^{6} + \frac{2016 x ^{5}}{5} + 168 x^{4} + 48 x^{3} + 9 x^{2} + x) + C

解答ステップ

\int 2 (2 x + 1)^{9} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (2 x + 1)^{9} d x

拡張

(2 x + 1)^{9} : 512 x^{9} + 2304 x^{8} + 4608 x^{7} + 5376 x^{6} + 4032 x^{5} + 2016 x^{4} + 672 x^{3} + 144 x^{2} + 18 x + 1

= 2 \cdot \int 512 x^{9} + 2304 x^{8} + 4608 x^{7} + 5376 x^{6} + 4032 x^{5} + 2016 x^{4} + 672 x^{3} + 144 x^{2} + 18 x + 1 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

= 2 (\int 512 x^{9} d x + \int 2304 x^{8} d x + \int 4608 x^{7} d x + \int 5376 x^{6} d x + \int 4032 x^{5} d x + \int 2016 x^{4} d x + \int 672 x^{3} d x + \int 144 x^{2} d x + \int 18 x d x + \int 1 d x)

\int 512 x^{9} d x = \frac{256 x ^{10}}{5}

\int 2304 x^{8} d x = 256 x^{9}

\int 4608 x^{7} d x = 576 x^{8}

\int 5376 x^{6} d x = 768 x^{7}

\int 4032 x^{5} d x = 672 x^{6}

\int 2016 x^{4} d x = \frac{2016 x ^{5}}{5}

\int 672 x^{3} d x = 168 x^{4}

\int 144 x^{2} d x = 48 x^{3}

\int 18 x d x = 9 x^{2}

\int 1 d x = x

= 2 (\frac{256 x ^{10}}{5} + 256 x^{9} + 576 x^{8} + 768 x^{7} + 672 x^{6} + \frac{2016 x ^{5}}{5} + 168 x^{4} + 48 x^{3} + 9 x^{2} + x)

定数を解答に追加する

= 2 (\frac{256 x ^{10}}{5} + 256 x^{9} + 576 x^{8} + 768 x^{7} + 672 x^{6} + \frac{2016 x ^{5}}{5} + 168 x^{4} + 48 x^{3} + 9 x^{2} + x) + C