integral of x^2(2x^3+5)^4

解

\int x^{2} (2 x^{3} + 5)^{4} d x

\frac{16 x ^{15}}{15} + \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} + \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3} + C

解答ステップ

\int x^{2} (2 x^{3} + 5)^{4} d x

拡張

x^{2} (2 x^{3} + 5)^{4} : 16 x^{14} + 160 x^{11} + 600 x^{8} + 1000 x^{5} + 625 x^{2}

= \int 16 x^{14} + 160 x^{11} + 600 x^{8} + 1000 x^{5} + 625 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 16 x^{14} d x + \int 160 x^{11} d x + \int 600 x^{8} d x + \int 1000 x^{5} d x + \int 625 x^{2} d x

\int 16 x^{14} d x = \frac{16 x ^{15}}{15}

\int 160 x^{11} d x = \frac{40 x ^{12}}{3}

\int 600 x^{8} d x = \frac{200 x ^{9}}{3}

\int 1000 x^{5} d x = \frac{500 x ^{6}}{3}

\int 625 x^{2} d x = \frac{625 x ^{3}}{3}

= \frac{16 x ^{15}}{15} + \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} + \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3}

定数を解答に追加する

= \frac{16 x ^{15}}{15} + \frac{40 x ^{12}}{3} + \frac{200 x ^{9}}{3} + \frac{500 x ^{6}}{3} + \frac{625 x ^{3}}{3} + C