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

解答

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

4 (\frac{4096 x ^{21}}{21} - \frac{1536 x ^{20}}{5} + \frac{3840 x ^{19}}{19} - \frac{640 x ^{18}}{9} + \frac{240 x ^{17}}{17} - \frac{3 x ^{16}}{2} + \frac{x ^{15}}{15}) + C

求解步骤

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

提出常数:

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

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

乘开

x^{2} (4 x^{3} - x^{2})^{6} : 4096 x^{20} - 6144 x^{19} + 3840 x^{18} - 1280 x^{17} + 240 x^{16} - 24 x^{15} + x^{14}

= 4 \cdot \int 4096 x^{20} - 6144 x^{19} + 3840 x^{18} - 1280 x^{17} + 240 x^{16} - 24 x^{15} + x^{14} 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 4096 x^{20} d x - \int 6144 x^{19} d x + \int 3840 x^{18} d x - \int 1280 x^{17} d x + \int 240 x^{16} d x - \int 24 x^{15} d x + \int x^{14} d x)

\int 4096 x^{20} d x = \frac{4096 x ^{21}}{21}

\int 6144 x^{19} d x = \frac{1536 x ^{20}}{5}

\int 3840 x^{18} d x = \frac{3840 x ^{19}}{19}

\int 1280 x^{17} d x = \frac{640 x ^{18}}{9}

\int 240 x^{16} d x = \frac{240 x ^{17}}{17}

\int 24 x^{15} d x = \frac{3 x ^{16}}{2}

\int x^{14} d x = \frac{x ^{15}}{15}

= 4 (\frac{4096 x ^{21}}{21} - \frac{1536 x ^{20}}{5} + \frac{3840 x ^{19}}{19} - \frac{640 x ^{18}}{9} + \frac{240 x ^{17}}{17} - \frac{3 x ^{16}}{2} + \frac{x ^{15}}{15})

解答补常数

= 4 (\frac{4096 x ^{21}}{21} - \frac{1536 x ^{20}}{5} + \frac{3840 x ^{19}}{19} - \frac{640 x ^{18}}{9} + \frac{240 x ^{17}}{17} - \frac{3 x ^{16}}{2} + \frac{x ^{15}}{15}) + C

y^{^{''}} + y^{^{'}} - 6 y = t e^{2 t}

\frac{\partial}{\partial x} (1 + \frac{2}{y})

\int_{0}^{34} (2 x - x^{2}) d x

x^{2} \frac{d ^{2} y}{d x ^{2}} - 7 x \frac{d y}{d x} + 16 y = 0

y^{^{'}} + 0.5 y = 2