integral of x(x+2)^6

解

\int x (x + 2)^{6} d x

\frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2} + C

解答ステップ

\int x (x + 2)^{6} d x

拡張

x (x + 2)^{6} : x^{7} + 12 x^{6} + 60 x^{5} + 160 x^{4} + 240 x^{3} + 192 x^{2} + 64 x

= \int x^{7} + 12 x^{6} + 60 x^{5} + 160 x^{4} + 240 x^{3} + 192 x^{2} + 64 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 x^{7} d x + \int 12 x^{6} d x + \int 60 x^{5} d x + \int 160 x^{4} d x + \int 240 x^{3} d x + \int 192 x^{2} d x + \int 64 x d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int 12 x^{6} d x = \frac{12 x ^{7}}{7}

\int 60 x^{5} d x = 10 x^{6}

\int 160 x^{4} d x = 32 x^{5}

\int 240 x^{3} d x = 60 x^{4}

\int 192 x^{2} d x = 64 x^{3}

\int 64 x d x = 32 x^{2}

= \frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2}

定数を解答に追加する

= \frac{x ^{8}}{8} + \frac{12 x ^{7}}{7} + 10 x^{6} + 32 x^{5} + 60 x^{4} + 64 x^{3} + 32 x^{2} + C