integral of (x+4)(2x+1)

解

\int (x + 4) (2 x + 1) d x

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

解答ステップ

\int (x + 4) (2 x + 1) d x

拡張

(x + 4) (2 x + 1) : 2 x^{2} + 9 x + 4

= \int 2 x^{2} + 9 x + 4 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 2 x^{2} d x + \int 9 x d x + \int 4 d x

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

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

\int 4 d x = 4 x

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

定数を解答に追加する

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