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

解答

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

2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2}) + C

求解步骤

\int 2 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

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

乘开

x (3 x - 2)^{6} : 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x

= 2 \cdot \int 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x 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

= 2 (\int 729 x^{7} d x - \int 2916 x^{6} d x + \int 4860 x^{5} d x - \int 4320 x^{4} d x + \int 2160 x^{3} d x - \int 576 x^{2} d x + \int 64 x d x)

\int 729 x^{7} d x = \frac{729 x ^{8}}{8}

\int 2916 x^{6} d x = \frac{2916 x ^{7}}{7}

\int 4860 x^{5} d x = 810 x^{6}

\int 4320 x^{4} d x = 864 x^{5}

\int 2160 x^{3} d x = 540 x^{4}

\int 576 x^{2} d x = 192 x^{3}

\int 64 x d x = 32 x^{2}

= 2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2})

解答补常数

= 2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2}) + C

s l o p e (5, 9), (4, 5)

t a y l or x^{\frac{1}{3}}, x = 8

x \to 0 lim (\frac{- x ^{3}}{x ^{2}})

t an g e n t f (x) = a x^{2}, at x = - 2

\int \frac{2 x ^{3} + 3 x ^{2} + x + 1}{2 x + 1} d x