integral of x(x-5)^2(x+1)

解

\int x (x - 5)^{2} (x + 1) d x

\frac{x ^{5}}{5} - \frac{9 x ^{4}}{4} + 5 x^{3} + \frac{25 x ^{2}}{2} + C

解答ステップ

\int x (x - 5)^{2} (x + 1) d x

拡張

x (x - 5)^{2} (x + 1) : x^{4} - 9 x^{3} + 15 x^{2} + 25 x

= \int x^{4} - 9 x^{3} + 15 x^{2} + 25 x d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{4} d x - \int 9 x^{3} d x + \int 15 x^{2} d x + \int 25 x d x

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

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

\int 15 x^{2} d x = 5 x^{3}

\int 25 x d x = \frac{25 x ^{2}}{2}

= \frac{x ^{5}}{5} - \frac{9 x ^{4}}{4} + 5 x^{3} + \frac{25 x ^{2}}{2}

定数を解答に追加する

= \frac{x ^{5}}{5} - \frac{9 x ^{4}}{4} + 5 x^{3} + \frac{25 x ^{2}}{2} + C