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

Lösung

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

- \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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Wende die Summenregel an:

\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)))

Setze in

u = x - 1

ein

= - \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))))

Vereinfache

- \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))))

Füge eine Konstante zur Lösung hinzu

= - \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

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