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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

\int 6 a x d x = 3 a x^{2}

\int 9 d x = 9 x

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

定数を解答に追加する

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