integral of 36x(2x^2-5)^8

解

\int 36 x (2 x^{2} - 5)^{8} d x

36 (\frac{128 x ^{18}}{9} - 320 x^{16} + 3200 x^{14} - \frac{56000 x ^{12}}{3} + 70000 x^{10} - 175000 x^{8} + \frac{875000 x ^{6}}{3} - 312500 x^{4} + \frac{390625 x ^{2}}{2}) + C

解答ステップ

\int 36 x (2 x^{2} - 5)^{8} d x

定数を除く:

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

= 36 \cdot \int x (2 x^{2} - 5)^{8} d x

拡張

x (2 x^{2} - 5)^{8} : 256 x^{17} - 5120 x^{15} + 44800 x^{13} - 224000 x^{11} + 700000 x^{9} - 1400000 x^{7} + 1750000 x^{5} - 1250000 x^{3} + 390625 x

= 36 \cdot \int 256 x^{17} - 5120 x^{15} + 44800 x^{13} - 224000 x^{11} + 700000 x^{9} - 1400000 x^{7} + 1750000 x^{5} - 1250000 x^{3} + 390625 x 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

= 36 (\int 256 x^{17} d x - \int 5120 x^{15} d x + \int 44800 x^{13} d x - \int 224000 x^{11} d x + \int 700000 x^{9} d x - \int 1400000 x^{7} d x + \int 1750000 x^{5} d x - \int 1250000 x^{3} d x + \int 390625 x d x)

\int 256 x^{17} d x = \frac{128 x ^{18}}{9}

\int 5120 x^{15} d x = 320 x^{16}

\int 44800 x^{13} d x = 3200 x^{14}

\int 224000 x^{11} d x = \frac{56000 x ^{12}}{3}

\int 700000 x^{9} d x = 70000 x^{10}

\int 1400000 x^{7} d x = 175000 x^{8}

\int 1750000 x^{5} d x = \frac{875000 x ^{6}}{3}

\int 1250000 x^{3} d x = 312500 x^{4}

\int 390625 x d x = \frac{390625 x ^{2}}{2}

= 36 (\frac{128 x ^{18}}{9} - 320 x^{16} + 3200 x^{14} - \frac{56000 x ^{12}}{3} + 70000 x^{10} - 175000 x^{8} + \frac{875000 x ^{6}}{3} - 312500 x^{4} + \frac{390625 x ^{2}}{2})

定数を解答に追加する

= 36 (\frac{128 x ^{18}}{9} - 320 x^{16} + 3200 x^{14} - \frac{56000 x ^{12}}{3} + 70000 x^{10} - 175000 x^{8} + \frac{875000 x ^{6}}{3} - 312500 x^{4} + \frac{390625 x ^{2}}{2}) + C