integral of (x^2-1)^4

解

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

\frac{x ^{9}}{9} - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} - \frac{4 x ^{3}}{3} + x + C

解答ステップ

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

拡張

(x^{2} - 1)^{4} : x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2} + 1

= \int x^{8} - 4 x^{6} + 6 x^{4} - 4 x^{2} + 1 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 x^{8} d x - \int 4 x^{6} d x + \int 6 x^{4} d x - \int 4 x^{2} d x + \int 1 d x

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

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

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

\int 1 d x = x

= \frac{x ^{9}}{9} - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} - \frac{4 x ^{3}}{3} + x

定数を解答に追加する

= \frac{x ^{9}}{9} - \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} - \frac{4 x ^{3}}{3} + x + C