integral of 4x^8(2x^9-1)^7

解答

\int 4 x^{8} (2 x^{9} - 1)^{7} d x

4 (\frac{16 x ^{72}}{9} - \frac{64 x ^{63}}{9} + \frac{112 x ^{54}}{9} - \frac{112 x ^{45}}{9} + \frac{70 x ^{36}}{9} - \frac{28 x ^{27}}{9} + \frac{7 x ^{18}}{9} - \frac{x ^{9}}{9}) + C

求解步骤

\int 4 x^{8} (2 x^{9} - 1)^{7} d x

提出常数:

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

= 4 \cdot \int x^{8} (2 x^{9} - 1)^{7} d x

乘开

x^{8} (2 x^{9} - 1)^{7} : 128 x^{71} - 448 x^{62} + 672 x^{53} - 560 x^{44} + 280 x^{35} - 84 x^{26} + 14 x^{17} - x^{8}

= 4 \cdot \int 128 x^{71} - 448 x^{62} + 672 x^{53} - 560 x^{44} + 280 x^{35} - 84 x^{26} + 14 x^{17} - x^{8} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int 128 x^{71} d x - \int 448 x^{62} d x + \int 672 x^{53} d x - \int 560 x^{44} d x + \int 280 x^{35} d x - \int 84 x^{26} d x + \int 14 x^{17} d x - \int x^{8} d x)

\int 128 x^{71} d x = \frac{16 x ^{72}}{9}

\int 448 x^{62} d x = \frac{64 x ^{63}}{9}

\int 672 x^{53} d x = \frac{112 x ^{54}}{9}

\int 560 x^{44} d x = \frac{112 x ^{45}}{9}

\int 280 x^{35} d x = \frac{70 x ^{36}}{9}

\int 84 x^{26} d x = \frac{28 x ^{27}}{9}

\int 14 x^{17} d x = \frac{7 x ^{18}}{9}

\int x^{8} d x = \frac{x ^{9}}{9}

= 4 (\frac{16 x ^{72}}{9} - \frac{64 x ^{63}}{9} + \frac{112 x ^{54}}{9} - \frac{112 x ^{45}}{9} + \frac{70 x ^{36}}{9} - \frac{28 x ^{27}}{9} + \frac{7 x ^{18}}{9} - \frac{x ^{9}}{9})

解答补常数

= 4 (\frac{16 x ^{72}}{9} - \frac{64 x ^{63}}{9} + \frac{112 x ^{54}}{9} - \frac{112 x ^{45}}{9} + \frac{70 x ^{36}}{9} - \frac{28 x ^{27}}{9} + \frac{7 x ^{18}}{9} - \frac{x ^{9}}{9}) + C

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