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

解

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

- \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} - 2 arcsin (\frac{1}{2} (x - 1)) - sin (2 arcsin (\frac{1}{2} (x - 1))) + C

解答ステップ

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

拡張

(x - 2) 3 + 2 x - x^{2} : x 3 + 2 x - x^{2} - 2 3 + 2 x - x^{2}

= \int x 3 + 2 x - x^{2} - 2 3 + 2 x - x^{2} 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 + 2 x - x^{2} d x - \int 2 3 + 2 x - x^{2} d x

\int x 3 + 2 x - x^{2} d x = - \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} + 2 arcsin (\frac{1}{2} (x - 1)) + sin (2 arcsin (\frac{1}{2} (x - 1)))

\int 2 3 + 2 x - x^{2} d x = 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1))))

= - \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} + 2 arcsin (\frac{1}{2} (x - 1)) + sin (2 arcsin (\frac{1}{2} (x - 1))) - 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1))))

簡素化

- \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} + 2 arcsin (\frac{1}{2} (x - 1)) + sin (2 arcsin (\frac{1}{2} (x - 1))) - 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1)))) : - \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} - 2 arcsin (\frac{1}{2} (x - 1)) - sin (2 arcsin (\frac{1}{2} (x - 1)))

= - \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} - 2 arcsin (\frac{1}{2} (x - 1)) - sin (2 arcsin (\frac{1}{2} (x - 1)))

定数を解答に追加する

= - \frac{1}{3} (- x^{2} + 2 x + 3)^{\frac{3}{2}} - 2 arcsin (\frac{1}{2} (x - 1)) - sin (2 arcsin (\frac{1}{2} (x - 1))) + C