integral of (7-2x)sqrt(9+16x-4x^2)

Lösung

\int (7 - 2 x) 9 + 16 x - 4 x^{2} d x

\frac{1}{6} (- 4 x^{2} + 16 x + 9)^{\frac{3}{2}} + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2)))) + C

Schritte zur Lösung

\int (7 - 2 x) 9 + 16 x - 4 x^{2} d x

Vervollständige das Quadrat

9 + 16 x - 4 x^{2} : - 4 (x - 2)^{2} + 25

= \int (7 - 2 x) - 4 (x - 2)^{2} + 25 d x

Wende U-Substitution an

= \int (- 2 u + 3) - 4 u^{2} + 25 d u

Multipliziere aus

(- 2 u + 3) - 4 u^{2} + 25 : - 2 u - 4 u^{2} + 25 + 3 - 4 u^{2} + 25

= \int - 2 u - 4 u^{2} + 25 + 3 - 4 u^{2} + 25 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 2 u - 4 u^{2} + 25 d u + \int 3 - 4 u^{2} + 25 d u

\int 2 u - 4 u^{2} + 25 d u = - \frac{1}{6} (- 4 u^{2} + 25)^{\frac{3}{2}}

\int 3 - 4 u^{2} + 25 d u = \frac{75}{4} (arcsin (\frac{2}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} u)))

= - (- \frac{1}{6} (- 4 u^{2} + 25)^{\frac{3}{2}}) + \frac{75}{4} (arcsin (\frac{2}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} u)))

Setze in

u = x - 2

ein

= - (- \frac{1}{6} (- 4 (x - 2)^{2} + 25)^{\frac{3}{2}}) + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2))))

Vereinfache

- (- \frac{1}{6} (- 4 (x - 2)^{2} + 25)^{\frac{3}{2}}) + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2)))) : \frac{1}{6} (- 4 x^{2} + 16 x + 9)^{\frac{3}{2}} + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2))))

= \frac{1}{6} (- 4 x^{2} + 16 x + 9)^{\frac{3}{2}} + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2))))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} (- 4 x^{2} + 16 x + 9)^{\frac{3}{2}} + \frac{75}{4} (arcsin (\frac{2}{5} (x - 2)) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - 2)))) + C

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