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

Lösung

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

- \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}} + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int - \frac{u}{4} d u

Entferne die Konstante:

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

= - \frac{1}{4} \cdot \int u d u

Wende die Potenzregel an

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Setze in

u = 4 - 2 x^{2} - x^{4}

ein

= - \frac{1}{4} \cdot \frac{2}{3} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}}

Vereinfache

- \frac{1}{4} \cdot \frac{2}{3} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}} : - \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}}

= - \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}} + C

x \to 2 lim (\frac{x - 3}{x - 2})

\int 2 t - 5 d t

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

\frac{d y}{d x} = \frac{( 2 sec ( y ) )}{( x - 4 ) ^{2}}

n = 2 \sum \infty \frac{1}{2 ^{n}}