integral of (7^{2/3}-x^{2/3})^3

解

\int (7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} d x

49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C

解答ステップ

\int (7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} d x

拡張

(7^{\frac{2}{3}} - x^{\frac{2}{3}})^{3} : 49 - 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} + 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} - x^{2}

= \int 49 - 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} + 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} - 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 49 d x - \int 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} d x + \int 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} d x - \int x^{2} d x

\int 49 d x = 49 x

\int 3 \cdot 7^{\frac{4}{3}} x^{\frac{2}{3}} d x = \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}}

\int 3 \cdot 7^{\frac{2}{3}} x^{\frac{4}{3}} d x = \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}}

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

= 49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3}

定数を解答に追加する

= 49 x - \frac{63 \cdot 7 ^{\frac{1}{3}}}{5} x^{\frac{5}{3}} + \frac{9}{7 ^{\frac{1}{3}}} x^{\frac{7}{3}} - \frac{x ^{3}}{3} + C