integral of 9(x^2+10x+5)(2x+10)

解

\int 9 (x^{2} + 10 x + 5) (2 x + 10) d x

9 (\frac{x ^{4}}{2} + 10 x^{3} + 55 x^{2} + 50 x) + C

解答ステップ

\int 9 (x^{2} + 10 x + 5) (2 x + 10) 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} + 10 x + 5) (2 x + 10) d x

拡張

(x^{2} + 10 x + 5) (2 x + 10) : 2 x^{3} + 30 x^{2} + 110 x + 50

= 9 \cdot \int 2 x^{3} + 30 x^{2} + 110 x + 50 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 30 x^{2} d x + \int 110 x d x + \int 50 d x)

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

\int 30 x^{2} d x = 10 x^{3}

\int 110 x d x = 55 x^{2}

\int 50 d x = 50 x

= 9 (\frac{x ^{4}}{2} + 10 x^{3} + 55 x^{2} + 50 x)

定数を解答に追加する

= 9 (\frac{x ^{4}}{2} + 10 x^{3} + 55 x^{2} + 50 x) + C