integral of (x^3)/(sqrt(1+x^2-x^4))

Lösung

\int \frac{x ^{3}}{1 + x ^{2} - x ^{4}} d x

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

Schritte zur Lösung

\int \frac{x ^{3}}{1 + x ^{2} - x ^{4}} d x

Wende U-Substitution an

= \int \frac{u}{2 u + 1 - u ^{2}} 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{u}{u + 1 - u ^{2}} d u

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Multipliziere aus

\frac{2 v + 1}{- 4 v ^{2} + 5} : \frac{2 v}{- 4 v ^{2} + 5} + \frac{1}{- 4 v ^{2} + 5}

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

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{2 v}{- 4 v ^{2} + 5} d v + \int \frac{1}{- 4 v ^{2} + 5} d v)

\int \frac{2 v}{- 4 v ^{2} + 5} d v = - \frac{1}{2} - 4 v^{2} + 5

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d y} (\frac{x y}{x ^{2} + y ^{2}})

d er i v a t i v e f (x) = \frac{2 - 5 x}{e ^{x}}

x \to \infty lim (5^{x})

d er i v a t i v e \frac{x}{2 + x}

\frac{\partial}{\partial x} (x cos (π y))