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

解

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

\frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x + C

解答ステップ

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

拡張

(4 x^{3} - 3 x + 2)^{2} (4 x^{2} - 1) : 64 x^{8} - 112 x^{6} + 64 x^{5} + 60 x^{4} - 64 x^{3} + 7 x^{2} + 12 x - 4

= \int 64 x^{8} - 112 x^{6} + 64 x^{5} + 60 x^{4} - 64 x^{3} + 7 x^{2} + 12 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 64 x^{8} d x - \int 112 x^{6} d x + \int 64 x^{5} d x + \int 60 x^{4} d x - \int 64 x^{3} d x + \int 7 x^{2} d x + \int 12 x d x - \int 4 d x

\int 64 x^{8} d x = \frac{64 x ^{9}}{9}

\int 112 x^{6} d x = 16 x^{7}

\int 64 x^{5} d x = \frac{32 x ^{6}}{3}

\int 60 x^{4} d x = 12 x^{5}

\int 64 x^{3} d x = 16 x^{4}

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 12 x d x = 6 x^{2}

\int 4 d x = 4 x

= \frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x

定数を解答に追加する

= \frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x + C