integral of 5x(2x+5)(x-3)

解

\int 5 x (2 x + 5) (x - 3) d x

5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2}) + C

解答ステップ

\int 5 x (2 x + 5) (x - 3) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int x (2 x + 5) (x - 3) d x

拡張

x (2 x + 5) (x - 3) : 2 x^{3} - x^{2} - 15 x

= 5 \cdot \int 2 x^{3} - x^{2} - 15 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

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

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

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

\int 15 x d x = \frac{15 x ^{2}}{2}

= 5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2})

定数を解答に追加する

= 5 (\frac{x ^{4}}{2} - \frac{x ^{3}}{3} - \frac{15 x ^{2}}{2}) + C