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

解答

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

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

求解步骤

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

乘开

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

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

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

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

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

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

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

解答补常数

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