integral of x(1+x)^4

解答

\int x (1 + x)^{4} d x

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

求解步骤

\int x (1 + x)^{4} d x

乘开

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

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

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

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

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

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

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

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

解答补常数

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

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