integral of (2x^2-1/3)^3

解

\int (2 x^{2} - \frac{1}{3})^{3} d x

\frac{8 x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{2 x ^{3}}{9} - \frac{1}{27} x + C

解答ステップ

\int (2 x^{2} - \frac{1}{3})^{3} d x

拡張

(2 x^{2} - \frac{1}{3})^{3} : 8 x^{6} - 4 x^{4} + \frac{2 x ^{2}}{3} - \frac{1}{27}

= \int 8 x^{6} - 4 x^{4} + \frac{2 x ^{2}}{3} - \frac{1}{27} 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 8 x^{6} d x - \int 4 x^{4} d x + \int \frac{2 x ^{2}}{3} d x - \int \frac{1}{27} d x

\int 8 x^{6} d x = \frac{8 x ^{7}}{7}

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

\int \frac{2 x ^{2}}{3} d x = \frac{2 x ^{3}}{9}

\int \frac{1}{27} d x = \frac{1}{27} x

= \frac{8 x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{2 x ^{3}}{9} - \frac{1}{27} x

定数を解答に追加する

= \frac{8 x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{2 x ^{3}}{9} - \frac{1}{27} x + C