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人気のある微分積分 >

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

解

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

解

(x + 1) (- x^{2} - 2 x + 3)^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1)))) + C

解答ステップ

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

(3 - 2 x - x^{2})^{3} = (3 - 2 x - x^{2})^{\frac{3}{2}}, （ (3 - 2 x - x^{2}) \geq 0

と仮定

）

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

平方完成する

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

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

u置換積分法を適用する

= \int (- u^{2} + 4)^{\frac{3}{2}} d u

部分積分を適用する

= u (- u^{2} + 4)^{\frac{3}{2}} - \int - 3 u^{2} - u^{2} + 4 d u

\int - 3 u^{2} - u^{2} + 4 d u = - 6 (arcsin (\frac{1}{2} u) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} u)))

= u (- u^{2} + 4)^{\frac{3}{2}} - (- 6 (arcsin (\frac{1}{2} u) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} u))))

代用を戻す

u = x + 1

= (x + 1) (- (x + 1)^{2} + 4)^{\frac{3}{2}} - (- 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1)))))

簡素化

(x + 1) (- (x + 1)^{2} + 4)^{\frac{3}{2}} - (- 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1))))) : (x + 1) (- x^{2} - 2 x + 3)^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1))))

= (x + 1) (- x^{2} - 2 x + 3)^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1))))

定数を解答に追加する

= (x + 1) (- x^{2} - 2 x + 3)^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} (x + 1)) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} (x + 1)))) + C

グラフ

人気の例

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