integral of 64x^3(4x^2-1)^3

解

\int 64 x^{3} (4 x^{2} - 1)^{3} d x

64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4}) + C

解答ステップ

\int 64 x^{3} (4 x^{2} - 1)^{3} d x

定数を除く:

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

= 64 \cdot \int x^{3} (4 x^{2} - 1)^{3} d x

拡張

x^{3} (4 x^{2} - 1)^{3} : 64 x^{9} - 48 x^{7} + 12 x^{5} - x^{3}

= 64 \cdot \int 64 x^{9} - 48 x^{7} + 12 x^{5} - x^{3} 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

= 64 (\int 64 x^{9} d x - \int 48 x^{7} d x + \int 12 x^{5} d x - \int x^{3} d x)

\int 64 x^{9} d x = \frac{32 x ^{10}}{5}

\int 48 x^{7} d x = 6 x^{8}

\int 12 x^{5} d x = 2 x^{6}

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

= 64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4})

定数を解答に追加する

= 64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4}) + C