integral of (2x+3)(2x-1)^{10}

解

\int (2 x + 3) (2 x - 1)^{10} d x

\frac{512 x ^{12}}{3} - \frac{7168 x ^{11}}{11} + 768 x^{10} + \frac{1280 x ^{9}}{3} - 2400 x^{8} + 3456 x^{7} - 2912 x^{6} + 1632 x^{5} - 630 x^{4} + \frac{500 x ^{3}}{3} - 29 x^{2} + 3 x + C

解答ステップ

\int (2 x + 3) (2 x - 1)^{10} d x

拡張

(2 x + 3) (2 x - 1)^{10} : 2048 x^{11} - 7168 x^{10} + 7680 x^{9} + 3840 x^{8} - 19200 x^{7} + 24192 x^{6} - 17472 x^{5} + 8160 x^{4} - 2520 x^{3} + 500 x^{2} - 58 x + 3

= \int 2048 x^{11} - 7168 x^{10} + 7680 x^{9} + 3840 x^{8} - 19200 x^{7} + 24192 x^{6} - 17472 x^{5} + 8160 x^{4} - 2520 x^{3} + 500 x^{2} - 58 x + 3 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 2048 x^{11} d x - \int 7168 x^{10} d x + \int 7680 x^{9} d x + \int 3840 x^{8} d x - \int 19200 x^{7} d x + \int 24192 x^{6} d x - \int 17472 x^{5} d x + \int 8160 x^{4} d x - \int 2520 x^{3} d x + \int 500 x^{2} d x - \int 58 x d x + \int 3 d x

\int 2048 x^{11} d x = \frac{512 x ^{12}}{3}

\int 7168 x^{10} d x = \frac{7168 x ^{11}}{11}

\int 7680 x^{9} d x = 768 x^{10}

\int 3840 x^{8} d x = \frac{1280 x ^{9}}{3}

\int 19200 x^{7} d x = 2400 x^{8}

\int 24192 x^{6} d x = 3456 x^{7}

\int 17472 x^{5} d x = 2912 x^{6}

\int 8160 x^{4} d x = 1632 x^{5}

\int 2520 x^{3} d x = 630 x^{4}

\int 500 x^{2} d x = \frac{500 x ^{3}}{3}

\int 58 x d x = 29 x^{2}

\int 3 d x = 3 x

= \frac{512 x ^{12}}{3} - \frac{7168 x ^{11}}{11} + 768 x^{10} + \frac{1280 x ^{9}}{3} - 2400 x^{8} + 3456 x^{7} - 2912 x^{6} + 1632 x^{5} - 630 x^{4} + \frac{500 x ^{3}}{3} - 29 x^{2} + 3 x

定数を解答に追加する

= \frac{512 x ^{12}}{3} - \frac{7168 x ^{11}}{11} + 768 x^{10} + \frac{1280 x ^{9}}{3} - 2400 x^{8} + 3456 x^{7} - 2912 x^{6} + 1632 x^{5} - 630 x^{4} + \frac{500 x ^{3}}{3} - 29 x^{2} + 3 x + C