integral of (x+1)(x+3)^2

解

\int (x + 1) (x + 3)^{2} d x

\frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x + C

解答ステップ

\int (x + 1) (x + 3)^{2} d x

拡張

(x + 1) (x + 3)^{2} : x^{3} + 7 x^{2} + 15 x + 9

= \int x^{3} + 7 x^{2} + 15 x + 9 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^{3} d x + \int 7 x^{2} d x + \int 15 x d x + \int 9 d x

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

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

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

\int 9 d x = 9 x

= \frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x

定数を解答に追加する

= \frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x + C