integral of x(2x+1)^5

解

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

\frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2} + C

解答ステップ

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

拡張

x (2 x + 1)^{5} : 32 x^{6} + 80 x^{5} + 80 x^{4} + 40 x^{3} + 10 x^{2} + x

= \int 32 x^{6} + 80 x^{5} + 80 x^{4} + 40 x^{3} + 10 x^{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 32 x^{6} d x + \int 80 x^{5} d x + \int 80 x^{4} d x + \int 40 x^{3} d x + \int 10 x^{2} d x + \int x d x

\int 32 x^{6} d x = \frac{32 x ^{7}}{7}

\int 80 x^{5} d x = \frac{40 x ^{6}}{3}

\int 80 x^{4} d x = 16 x^{5}

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

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

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

= \frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2}

定数を解答に追加する

= \frac{32 x ^{7}}{7} + \frac{40 x ^{6}}{3} + 16 x^{5} + 10 x^{4} + \frac{10 x ^{3}}{3} + \frac{x ^{2}}{2} + C