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

解

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

\frac{125 x ^{10}}{2} + \frac{125 x ^{8}}{3} + \frac{100 x ^{6}}{9} + \frac{40 x ^{4}}{27} + \frac{8 x ^{2}}{81} + C

解答ステップ

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

拡張

x (5 x^{2} + \frac{2}{3})^{4} : 625 x^{9} + \frac{1000 x ^{7}}{3} + \frac{200 x ^{5}}{3} + \frac{160 x ^{3}}{27} + \frac{16 x}{81}

= \int 625 x^{9} + \frac{1000 x ^{7}}{3} + \frac{200 x ^{5}}{3} + \frac{160 x ^{3}}{27} + \frac{16 x}{81} 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 625 x^{9} d x + \int \frac{1000 x ^{7}}{3} d x + \int \frac{200 x ^{5}}{3} d x + \int \frac{160 x ^{3}}{27} d x + \int \frac{16 x}{81} d x

\int 625 x^{9} d x = \frac{125 x ^{10}}{2}

\int \frac{1000 x ^{7}}{3} d x = \frac{125 x ^{8}}{3}

\int \frac{200 x ^{5}}{3} d x = \frac{100 x ^{6}}{9}

\int \frac{160 x ^{3}}{27} d x = \frac{40 x ^{4}}{27}

\int \frac{16 x}{81} d x = \frac{8 x ^{2}}{81}

= \frac{125 x ^{10}}{2} + \frac{125 x ^{8}}{3} + \frac{100 x ^{6}}{9} + \frac{40 x ^{4}}{27} + \frac{8 x ^{2}}{81}

定数を解答に追加する

= \frac{125 x ^{10}}{2} + \frac{125 x ^{8}}{3} + \frac{100 x ^{6}}{9} + \frac{40 x ^{4}}{27} + \frac{8 x ^{2}}{81} + C