integral of x^2(x-1)^4

解答

\int x^{2} (x - 1)^{4} d x

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

求解步骤

\int x^{2} (x - 1)^{4} d x

乘开

x^{2} (x - 1)^{4} : x^{6} - 4 x^{5} + 6 x^{4} - 4 x^{3} + x^{2}

= \int x^{6} - 4 x^{5} + 6 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 x^{6} d x - \int 4 x^{5} d x + \int 6 x^{4} d x - \int 4 x^{3} d x + \int x^{2} d x

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

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

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

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

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

解答补常数

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

\int 5 x^{4} - 3 x^{2} + 4 d x

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