integral of (4x+1)(2x^2+x-3)^4

解

\int (4 x + 1) (2 x^{2} + x - 3)^{4} d x

\frac{32 x ^{10}}{5} + 16 x^{9} - 32 x^{8} - 88 x^{7} + 74 x^{6} + \frac{961 x ^{5}}{5} - 111 x^{4} - 198 x^{3} + 108 x^{2} + 81 x + C

解答ステップ

\int (4 x + 1) (2 x^{2} + x - 3)^{4} d x

拡張

(4 x + 1) (2 x^{2} + x - 3)^{4} : 64 x^{9} + 144 x^{8} - 256 x^{7} - 616 x^{6} + 444 x^{5} + 961 x^{4} - 444 x^{3} - 594 x^{2} + 216 x + 81

= \int 64 x^{9} + 144 x^{8} - 256 x^{7} - 616 x^{6} + 444 x^{5} + 961 x^{4} - 444 x^{3} - 594 x^{2} + 216 x + 81 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 64 x^{9} d x + \int 144 x^{8} d x - \int 256 x^{7} d x - \int 616 x^{6} d x + \int 444 x^{5} d x + \int 961 x^{4} d x - \int 444 x^{3} d x - \int 594 x^{2} d x + \int 216 x d x + \int 81 d x

\int 64 x^{9} d x = \frac{32 x ^{10}}{5}

\int 144 x^{8} d x = 16 x^{9}

\int 256 x^{7} d x = 32 x^{8}

\int 616 x^{6} d x = 88 x^{7}

\int 444 x^{5} d x = 74 x^{6}

\int 961 x^{4} d x = \frac{961 x ^{5}}{5}

\int 444 x^{3} d x = 111 x^{4}

\int 594 x^{2} d x = 198 x^{3}

\int 216 x d x = 108 x^{2}

\int 81 d x = 81 x

= \frac{32 x ^{10}}{5} + 16 x^{9} - 32 x^{8} - 88 x^{7} + 74 x^{6} + \frac{961 x ^{5}}{5} - 111 x^{4} - 198 x^{3} + 108 x^{2} + 81 x

定数を解答に追加する

= \frac{32 x ^{10}}{5} + 16 x^{9} - 32 x^{8} - 88 x^{7} + 74 x^{6} + \frac{961 x ^{5}}{5} - 111 x^{4} - 198 x^{3} + 108 x^{2} + 81 x + C