integral of (4x+2)(x^2+x-1)^5

解

\int (4 x + 2) (x^{2} + x - 1)^{5} d x

\frac{x ^{12}}{3} + 2 x^{11} + 3 x^{10} - \frac{10 x ^{9}}{3} - 10 x^{8} + 2 x^{7} + \frac{41 x ^{6}}{3} - 2 x^{5} - 10 x^{4} + \frac{10 x ^{3}}{3} + 3 x^{2} - 2 x + C

解答ステップ

\int (4 x + 2) (x^{2} + x - 1)^{5} d x

拡張

(4 x + 2) (x^{2} + x - 1)^{5} : 4 x^{11} + 22 x^{10} + 30 x^{9} - 30 x^{8} - 80 x^{7} + 14 x^{6} + 82 x^{5} - 10 x^{4} - 40 x^{3} + 10 x^{2} + 6 x - 2

= \int 4 x^{11} + 22 x^{10} + 30 x^{9} - 30 x^{8} - 80 x^{7} + 14 x^{6} + 82 x^{5} - 10 x^{4} - 40 x^{3} + 10 x^{2} + 6 x - 2 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 4 x^{11} d x + \int 22 x^{10} d x + \int 30 x^{9} d x - \int 30 x^{8} d x - \int 80 x^{7} d x + \int 14 x^{6} d x + \int 82 x^{5} d x - \int 10 x^{4} d x - \int 40 x^{3} d x + \int 10 x^{2} d x + \int 6 x d x - \int 2 d x

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

\int 22 x^{10} d x = 2 x^{11}

\int 30 x^{9} d x = 3 x^{10}

\int 30 x^{8} d x = \frac{10 x ^{9}}{3}

\int 80 x^{7} d x = 10 x^{8}

\int 14 x^{6} d x = 2 x^{7}

\int 82 x^{5} d x = \frac{41 x ^{6}}{3}

\int 10 x^{4} d x = 2 x^{5}

\int 40 x^{3} d x = 10 x^{4}

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

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

\int 2 d x = 2 x

= \frac{x ^{12}}{3} + 2 x^{11} + 3 x^{10} - \frac{10 x ^{9}}{3} - 10 x^{8} + 2 x^{7} + \frac{41 x ^{6}}{3} - 2 x^{5} - 10 x^{4} + \frac{10 x ^{3}}{3} + 3 x^{2} - 2 x

定数を解答に追加する

= \frac{x ^{12}}{3} + 2 x^{11} + 3 x^{10} - \frac{10 x ^{9}}{3} - 10 x^{8} + 2 x^{7} + \frac{41 x ^{6}}{3} - 2 x^{5} - 10 x^{4} + \frac{10 x ^{3}}{3} + 3 x^{2} - 2 x + C