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人気のある微分積分 >

integral of xsqrt(8+2x-x^2)

解

\int x 8 + 2 x - x^{2} d x

解

- \frac{1}{3} (- x^{2} + 2 x + 8)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1)))) + C

解答ステップ

\int x 8 + 2 x - x^{2} d x

平方完成する

8 + 2 x - x^{2} : - (x - 1)^{2} + 9

= \int x - (x - 1)^{2} + 9 d x

u置換積分法を適用する

= \int (u + 1) - u^{2} + 9 d u

拡張

(u + 1) - u^{2} + 9 : u - u^{2} + 9 + - u^{2} + 9

= \int u - u^{2} + 9 + - u^{2} + 9 d u

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 u - u^{2} + 9 d u + \int - u^{2} + 9 d u

\int u - u^{2} + 9 d u = - \frac{1}{3} (- u^{2} + 9)^{\frac{3}{2}}

\int - u^{2} + 9 d u = \frac{9}{2} (arcsin (\frac{1}{3} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} u)))

= - \frac{1}{3} (- u^{2} + 9)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} u)))

代用を戻す

u = x - 1

= - \frac{1}{3} (- (x - 1)^{2} + 9)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1))))

簡素化

- \frac{1}{3} (- (x - 1)^{2} + 9)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1)))) : - \frac{1}{3} (- x^{2} + 2 x + 8)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1))))

= - \frac{1}{3} (- x^{2} + 2 x + 8)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1))))

定数を解答に追加する

= - \frac{1}{3} (- x^{2} + 2 x + 8)^{\frac{3}{2}} + \frac{9}{2} (arcsin (\frac{1}{3} (x - 1)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 1)))) + C

グラフ

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