integral of 6x^3(-7-3x^4)^2

解

\int 6 x^{3} (- 7 - 3 x^{4})^{2} d x

6 (\frac{49 x ^{4}}{4} + \frac{21 x ^{8}}{4} + \frac{3 x ^{12}}{4}) + C

解答ステップ

\int 6 x^{3} (- 7 - 3 x^{4})^{2} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x^{3} (- 7 - 3 x^{4})^{2} d x

- 7 - 3 x^{4} = 7 (- 1 - \frac{3 x ^{4}}{7})

= 6 \cdot \int x^{3} (7 (- 1 - \frac{3 x ^{4}}{7}))^{2} d x

簡素化

x^{3} (7 (- 1 - \frac{3 x ^{4}}{7}))^{2} : 49 x^{3} + 42 x^{7} + 9 x^{11}

= 6 \cdot \int 49 x^{3} + 42 x^{7} + 9 x^{11} 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

= 6 (\int 49 x^{3} d x + \int 42 x^{7} d x + \int 9 x^{11} d x)

\int 49 x^{3} d x = \frac{49 x ^{4}}{4}

\int 42 x^{7} d x = \frac{21 x ^{8}}{4}

\int 9 x^{11} d x = \frac{3 x ^{12}}{4}

= 6 (\frac{49 x ^{4}}{4} + \frac{21 x ^{8}}{4} + \frac{3 x ^{12}}{4})

定数を解答に追加する

= 6 (\frac{49 x ^{4}}{4} + \frac{21 x ^{8}}{4} + \frac{3 x ^{12}}{4}) + C