integral of (x^2+4)^7

解答

\int (x^{2} + 4)^{7} d x

\frac{x ^{15}}{15} + \frac{28 x ^{13}}{13} + \frac{336 x ^{11}}{11} + \frac{2240 x ^{9}}{9} + 1280 x^{7} + \frac{21504 x ^{5}}{5} + \frac{28672 x ^{3}}{3} + 16384 x + C

求解步骤

\int (x^{2} + 4)^{7} d x

乘开

(x^{2} + 4)^{7} : x^{14} + 28 x^{12} + 336 x^{10} + 2240 x^{8} + 8960 x^{6} + 21504 x^{4} + 28672 x^{2} + 16384

= \int x^{14} + 28 x^{12} + 336 x^{10} + 2240 x^{8} + 8960 x^{6} + 21504 x^{4} + 28672 x^{2} + 16384 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

= \int x^{14} d x + \int 28 x^{12} d x + \int 336 x^{10} d x + \int 2240 x^{8} d x + \int 8960 x^{6} d x + \int 21504 x^{4} d x + \int 28672 x^{2} d x + \int 16384 d x

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

\int 28 x^{12} d x = \frac{28 x ^{13}}{13}

\int 336 x^{10} d x = \frac{336 x ^{11}}{11}

\int 2240 x^{8} d x = \frac{2240 x ^{9}}{9}

\int 8960 x^{6} d x = 1280 x^{7}

\int 21504 x^{4} d x = \frac{21504 x ^{5}}{5}

\int 28672 x^{2} d x = \frac{28672 x ^{3}}{3}

\int 16384 d x = 16384 x

= \frac{x ^{15}}{15} + \frac{28 x ^{13}}{13} + \frac{336 x ^{11}}{11} + \frac{2240 x ^{9}}{9} + 1280 x^{7} + \frac{21504 x ^{5}}{5} + \frac{28672 x ^{3}}{3} + 16384 x

解答补常数

= \frac{x ^{15}}{15} + \frac{28 x ^{13}}{13} + \frac{336 x ^{11}}{11} + \frac{2240 x ^{9}}{9} + 1280 x^{7} + \frac{21504 x ^{5}}{5} + \frac{28672 x ^{3}}{3} + 16384 x + C

d er i v a t i v e f (x) = x^{2} arctan (3 x)

\int y^{2} e^{x y} d x

\frac{\partial}{\partial y} (y^{2} cos (y))

x \to 3 lim (\frac{x ^{2} + 1}{x - 2})

d er i v a t i v e f (x) = (x^{3} + 5) e^{x}