integral of x^2(x+1)

解答

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

\frac{x ^{4}}{4} + \frac{x ^{3}}{3} + C

求解步骤

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

乘开

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

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

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

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

= \frac{x ^{4}}{4} + \frac{x ^{3}}{3}

解答补常数

= \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + C