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

Lösung

\int x^{2} + 2 x^{4} d x

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

Schritte zur Lösung

\int x^{2} + 2 x^{4} d x

Faktorisiere

x^{2} + 2 x^{4} : x^{2} 1 + 2 x^{2}

= \int x^{2} 1 + 2 x^{2} d x

x^{2} = (x),

angenommen

x \geq 0

= \int x 1 + 2 x^{2} 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 = 1 + 2 x^{2}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (\frac{sin ^{5} ( x )}{5} - \frac{sin ^{7} ( x )}{7})

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

d er i v a t i v e y = (2 x + 1) (x + 5)^{3}

n = 0 \sum \infty (- 1)^{n} (2^{n})

d er i v a t i v e y = 30 e^{- 0.0000323 (1600)}