integral of 20x^4(4x^5-1)^4

解

\int 20 x^{4} (4 x^{5} - 1)^{4} d x

20 (\frac{256 x ^{25}}{25} - \frac{64 x ^{20}}{5} + \frac{32 x ^{15}}{5} - \frac{8 x ^{10}}{5} + \frac{x ^{5}}{5}) + C

解答ステップ

\int 20 x^{4} (4 x^{5} - 1)^{4} d x

定数を除く:

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

= 20 \cdot \int x^{4} (4 x^{5} - 1)^{4} d x

拡張

x^{4} (4 x^{5} - 1)^{4} : 256 x^{24} - 256 x^{19} + 96 x^{14} - 16 x^{9} + x^{4}

= 20 \cdot \int 256 x^{24} - 256 x^{19} + 96 x^{14} - 16 x^{9} + x^{4} 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

= 20 (\int 256 x^{24} d x - \int 256 x^{19} d x + \int 96 x^{14} d x - \int 16 x^{9} d x + \int x^{4} d x)

\int 256 x^{24} d x = \frac{256 x ^{25}}{25}

\int 256 x^{19} d x = \frac{64 x ^{20}}{5}

\int 96 x^{14} d x = \frac{32 x ^{15}}{5}

\int 16 x^{9} d x = \frac{8 x ^{10}}{5}

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

= 20 (\frac{256 x ^{25}}{25} - \frac{64 x ^{20}}{5} + \frac{32 x ^{15}}{5} - \frac{8 x ^{10}}{5} + \frac{x ^{5}}{5})

定数を解答に追加する

= 20 (\frac{256 x ^{25}}{25} - \frac{64 x ^{20}}{5} + \frac{32 x ^{15}}{5} - \frac{8 x ^{10}}{5} + \frac{x ^{5}}{5}) + C