integral of (x+1)(x+2)^6

解答

\int (x + 1) (x + 2)^{6} d x

\frac{x ^{8}}{8} + \frac{13 x ^{7}}{7} + 12 x^{6} + 44 x^{5} + 100 x^{4} + 144 x^{3} + 128 x^{2} + 64 x + C

求解步骤

\int (x + 1) (x + 2)^{6} d x

乘开

(x + 1) (x + 2)^{6} : x^{7} + 13 x^{6} + 72 x^{5} + 220 x^{4} + 400 x^{3} + 432 x^{2} + 256 x + 64

= \int x^{7} + 13 x^{6} + 72 x^{5} + 220 x^{4} + 400 x^{3} + 432 x^{2} + 256 x + 64 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^{7} d x + \int 13 x^{6} d x + \int 72 x^{5} d x + \int 220 x^{4} d x + \int 400 x^{3} d x + \int 432 x^{2} d x + \int 256 x d x + \int 64 d x

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

\int 13 x^{6} d x = \frac{13 x ^{7}}{7}

\int 72 x^{5} d x = 12 x^{6}

\int 220 x^{4} d x = 44 x^{5}

\int 400 x^{3} d x = 100 x^{4}

\int 432 x^{2} d x = 144 x^{3}

\int 256 x d x = 128 x^{2}

\int 64 d x = 64 x

= \frac{x ^{8}}{8} + \frac{13 x ^{7}}{7} + 12 x^{6} + 44 x^{5} + 100 x^{4} + 144 x^{3} + 128 x^{2} + 64 x

解答补常数

= \frac{x ^{8}}{8} + \frac{13 x ^{7}}{7} + 12 x^{6} + 44 x^{5} + 100 x^{4} + 144 x^{3} + 128 x^{2} + 64 x + C

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