integral of (6x^5+2)(2x^6+4x)^4

解

\int (6 x^{5} + 2) (2 x^{6} + 4 x)^{4} d x

\frac{16 x ^{30}}{5} + 32 x^{25} + 128 x^{20} + 256 x^{15} + 256 x^{10} + \frac{512 x ^{5}}{5} + C

解答ステップ

\int (6 x^{5} + 2) (2 x^{6} + 4 x)^{4} d x

拡張

(6 x^{5} + 2) (2 x^{6} + 4 x)^{4} : 96 x^{29} + 800 x^{24} + 2560 x^{19} + 3840 x^{14} + 2560 x^{9} + 512 x^{4}

= \int 96 x^{29} + 800 x^{24} + 2560 x^{19} + 3840 x^{14} + 2560 x^{9} + 512 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 96 x^{29} d x + \int 800 x^{24} d x + \int 2560 x^{19} d x + \int 3840 x^{14} d x + \int 2560 x^{9} d x + \int 512 x^{4} d x

\int 96 x^{29} d x = \frac{16 x ^{30}}{5}

\int 800 x^{24} d x = 32 x^{25}

\int 2560 x^{19} d x = 128 x^{20}

\int 3840 x^{14} d x = 256 x^{15}

\int 2560 x^{9} d x = 256 x^{10}

\int 512 x^{4} d x = \frac{512 x ^{5}}{5}

= \frac{16 x ^{30}}{5} + 32 x^{25} + 128 x^{20} + 256 x^{15} + 256 x^{10} + \frac{512 x ^{5}}{5}

定数を解答に追加する

= \frac{16 x ^{30}}{5} + 32 x^{25} + 128 x^{20} + 256 x^{15} + 256 x^{10} + \frac{512 x ^{5}}{5} + C