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

解

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

\frac{x ^{6}}{6} - \frac{3 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 9 x + C

解答ステップ

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

拡張

(x^{2} - 2 x + 3)^{2} (x + 1) : x^{5} - 3 x^{4} + 6 x^{3} - 2 x^{2} - 3 x + 9

= \int x^{5} - 3 x^{4} + 6 x^{3} - 2 x^{2} - 3 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^{5} d x - \int 3 x^{4} d x + \int 6 x^{3} d x - \int 2 x^{2} d x - \int 3 x d x + \int 9 d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

\int 9 d x = 9 x

= \frac{x ^{6}}{6} - \frac{3 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 9 x

定数を解答に追加する

= \frac{x ^{6}}{6} - \frac{3 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 9 x + C