integral of x^2(x^3+1)^{1/4}

Lösung

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

\frac{4}{15} (x^{3} + 1)^{\frac{5}{4}} + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u ^{\frac{1}{4}}}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int u^{\frac{1}{4}} d u

Wende die Potenzregel an

= \frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}}

Setze in

u = x^{3} + 1

ein

= \frac{1}{3} \cdot \frac{4}{5} (x^{3} + 1)^{\frac{5}{4}}

Vereinfache

\frac{1}{3} \cdot \frac{4}{5} (x^{3} + 1)^{\frac{5}{4}} : \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}}

= \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}}

Füge eine Konstante zur Lösung hinzu

= \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}} + C

\frac{\partial}{\partial t} (2 cos (t))

f (x) = 2 arctan (x)

\frac{d}{d y} (\frac{1}{y})

\int - cos^{3} (x) d x

\frac{d}{d x} (sin (3) x^{2} cos (2 x))