integral of (3x^2-2x+1)(x^3-x^2+x)^2

解答

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{2} d x

\frac{x ^{9}}{3} - x^{8} + 2 x^{7} - \frac{7 x ^{6}}{3} + 2 x^{5} - x^{4} + \frac{x ^{3}}{3} + C

求解步骤

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{2} d x

乘开

(3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{2} : 3 x^{8} - 8 x^{7} + 14 x^{6} - 14 x^{5} + 10 x^{4} - 4 x^{3} + x^{2}

= \int 3 x^{8} - 8 x^{7} + 14 x^{6} - 14 x^{5} + 10 x^{4} - 4 x^{3} + x^{2} 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 3 x^{8} d x - \int 8 x^{7} d x + \int 14 x^{6} d x - \int 14 x^{5} d x + \int 10 x^{4} d x - \int 4 x^{3} d x + \int x^{2} d x

\int 3 x^{8} d x = \frac{x ^{9}}{3}

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

\int 14 x^{6} d x = 2 x^{7}

\int 14 x^{5} d x = \frac{7 x ^{6}}{3}

\int 10 x^{4} d x = 2 x^{5}

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

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

= \frac{x ^{9}}{3} - x^{8} + 2 x^{7} - \frac{7 x ^{6}}{3} + 2 x^{5} - x^{4} + \frac{x ^{3}}{3}

解答补常数

= \frac{x ^{9}}{3} - x^{8} + 2 x^{7} - \frac{7 x ^{6}}{3} + 2 x^{5} - x^{4} + \frac{x ^{3}}{3} + C

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