integral of (x^2)/(sqrt(1+x-x^2))

Lösung

\int \frac{x ^{2}}{1 + x - x ^{2}} d x

\frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1) + C

Schritte zur Lösung

\int \frac{x ^{2}}{1 + x - x ^{2}} d x

Vervollständige das Quadrat

1 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{5}{4}

= \int \frac{x ^{2}}{- ( x - \frac{1}{2} ) ^{2} + \frac{5}{4}} d x

Wende U-Substitution an

= \int \frac{( 2 u + 1 ) ^{2}}{2 - 4 u ^{2} + 5} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 5} d u

Multipliziere aus

\frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 5} : \frac{4 u ^{2}}{- 4 u ^{2} + 5} + \frac{4 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5}

= \frac{1}{2} \cdot \int \frac{4 u ^{2}}{- 4 u ^{2} + 5} + \frac{4 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5} 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

= \frac{1}{2} (\int \frac{4 u ^{2}}{- 4 u ^{2} + 5} d u + \int \frac{4 u}{- 4 u ^{2} + 5} d u + \int \frac{1}{- 4 u ^{2} + 5} d u)

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

\int \frac{4 u}{- 4 u ^{2} + 5} d u = - - 4 u^{2} + 5

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

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

Setze in

u = x - \frac{1}{2}

ein

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

Vereinfache

\frac{1}{2} \frac{5}{4} (arcsin (\frac{2}{5} (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 5 + \frac{1}{2} arcsin (\frac{2}{5} (x - \frac{1}{2})) : \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1)

= \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1) + C