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

解

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

\frac{x ^{12}}{4} - x^{11} + \frac{3 x ^{10}}{2} - \frac{11 x ^{8}}{4} + 3 x^{7} + \frac{x ^{6}}{2} - 3 x^{5} + \frac{3 x ^{4}}{2} + x^{3} - x^{2} + C

解答ステップ

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

拡張

(3 x^{2} - 2 x) (x^{3} - x^{2} + 1)^{3} : 3 x^{11} - 11 x^{10} + 15 x^{9} - 22 x^{7} + 21 x^{6} + 3 x^{5} - 15 x^{4} + 6 x^{3} + 3 x^{2} - 2 x

= \int 3 x^{11} - 11 x^{10} + 15 x^{9} - 22 x^{7} + 21 x^{6} + 3 x^{5} - 15 x^{4} + 6 x^{3} + 3 x^{2} - 2 x 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 3 x^{11} d x - \int 11 x^{10} d x + \int 15 x^{9} d x - \int 22 x^{7} d x + \int 21 x^{6} d x + \int 3 x^{5} d x - \int 15 x^{4} d x + \int 6 x^{3} d x + \int 3 x^{2} d x - \int 2 x d x

\int 3 x^{11} d x = \frac{x ^{12}}{4}

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

\int 15 x^{9} d x = \frac{3 x ^{10}}{2}

\int 22 x^{7} d x = \frac{11 x ^{8}}{4}

\int 21 x^{6} d x = 3 x^{7}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 15 x^{4} d x = 3 x^{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

= \frac{x ^{12}}{4} - x^{11} + \frac{3 x ^{10}}{2} - \frac{11 x ^{8}}{4} + 3 x^{7} + \frac{x ^{6}}{2} - 3 x^{5} + \frac{3 x ^{4}}{2} + x^{3} - x^{2}

定数を解答に追加する

= \frac{x ^{12}}{4} - x^{11} + \frac{3 x ^{10}}{2} - \frac{11 x ^{8}}{4} + 3 x^{7} + \frac{x ^{6}}{2} - 3 x^{5} + \frac{3 x ^{4}}{2} + x^{3} - x^{2} + C