integral of (2x^2-4x+4)/(sqrt(3+2x-x^2))

Lösung

\int \frac{2 x ^{2} - 4 x + 4}{3 + 2 x - x ^{2}} d x

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

Schritte zur Lösung

\int \frac{2 x ^{2} - 4 x + 4}{3 + 2 x - x ^{2}} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{2 u ^{2} + 2}{- u ^{2} + 4} : \frac{2 u ^{2}}{- u ^{2} + 4} + \frac{2}{- u ^{2} + 4}

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

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

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

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

Setze in

u = x - 1

ein

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

Vereinfache

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

= 6 arcsin (\frac{1}{2} (x - 1)) - 2 sin (2 arcsin (\frac{1}{2} (x - 1)))

Füge eine Konstante zur Lösung hinzu

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

y^{^{'}} = y - y^{2}

\frac{d y}{d x} = 10 - y

x \to - 5 + lim (\frac{x + 10}{x + 5})

\int x^{2} e^{16 x} d x

\frac{d}{d x} (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})