integral of x^5\sqrt[3]{x^3+1}

Lösung

\int x^{5} 3 x^{3} + 1 d x

\frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}}) + C

Schritte zur Lösung

\int x^{5} 3 x^{3} + 1 d x

Wende U-Substitution an

= \int \frac{( u - 1 ) 3 u}{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 - 1) 3 u d u

Multipliziere aus

(u - 1) 3 u : u^{\frac{4}{3}} - 3 u

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

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

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

= \frac{1}{3} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Setze in

u = x^{3} + 1

ein

= \frac{1}{3} (\frac{3}{7} (x^{3} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}})

Vereinfache

\frac{1}{3} (\frac{3}{7} (x^{3} + 1)^{\frac{7}{3}} - \frac{3}{4} (x^{3} + 1)^{\frac{4}{3}}) : \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}})

= \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{28} (4 (x^{3} + 1)^{\frac{7}{3}} - 7 (x^{3} + 1)^{\frac{4}{3}}) + C

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

in v erse l a pl a ce \frac{1}{( s + 3 ) ^{3}}

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

d er i v a t i v e g (t) = 4 t^{- \frac{3}{8}}

\int \frac{1}{x ^{2} x ^{2} - 36} d x