integral of 5x\sqrt[3]{(6x-1)^5}

Lösung

\int 5 x 3 (6 x - 1)^{5} d x

\frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}} + C

Schritte zur Lösung

\int 5 x 3 (6 x - 1)^{5} d x

Entferne die Konstante:

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

= 5 \cdot \int x 3 (6 x - 1)^{5} d x

Wende U-Substitution an

= 5 \cdot \int \frac{u ^{\frac{5}{3}} ( u + 1 )}{36} d u

Entferne die Konstante:

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

= 5 \cdot \frac{1}{36} \cdot \int u^{\frac{5}{3}} (u + 1) d u

Multipliziere aus

u^{\frac{5}{3}} (u + 1) : u^{\frac{8}{3}} + u^{\frac{5}{3}}

= 5 \cdot \frac{1}{36} \cdot \int u^{\frac{8}{3}} + u^{\frac{5}{3}} 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

= 5 \cdot \frac{1}{36} (\int u^{\frac{8}{3}} d u + \int u^{\frac{5}{3}} d u)

\int u^{\frac{8}{3}} d u = \frac{3}{11} u^{\frac{11}{3}}

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

= 5 \cdot \frac{1}{36} (\frac{3}{11} u^{\frac{11}{3}} + \frac{3}{8} u^{\frac{8}{3}})

Setze in

u = 6 x - 1

ein

= 5 \cdot \frac{1}{36} (\frac{3}{11} (6 x - 1)^{\frac{11}{3}} + \frac{3}{8} (6 x - 1)^{\frac{8}{3}})

Vereinfache

5 \cdot \frac{1}{36} (\frac{3}{11} (6 x - 1)^{\frac{11}{3}} + \frac{3}{8} (6 x - 1)^{\frac{8}{3}}) : \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}}

= \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}}

Füge eine Konstante zur Lösung hinzu

= \frac{5}{352} (16 x + 1) (6 x - 1)^{\frac{8}{3}} + C

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