integral of 9(x^2+3x+5)(2x+3)

解

\int 9 (x^{2} + 3 x + 5) (2 x + 3) d x

9 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{19 x ^{2}}{2} + 15 x) + C

解答ステップ

\int 9 (x^{2} + 3 x + 5) (2 x + 3) d x

定数を除く:

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

= 9 \cdot \int (x^{2} + 3 x + 5) (2 x + 3) d x

拡張

(x^{2} + 3 x + 5) (2 x + 3) : 2 x^{3} + 9 x^{2} + 19 x + 15

= 9 \cdot \int 2 x^{3} + 9 x^{2} + 19 x + 15 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

= 9 (\int 2 x^{3} d x + \int 9 x^{2} d x + \int 19 x d x + \int 15 d x)

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

\int 9 x^{2} d x = 3 x^{3}

\int 19 x d x = \frac{19 x ^{2}}{2}

\int 15 d x = 15 x

= 9 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{19 x ^{2}}{2} + 15 x)

定数を解答に追加する

= 9 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{19 x ^{2}}{2} + 15 x) + C