integral of (2x-3)\sqrt[3]{(2x^2-6x+7)}

Lösung

\int (2 x - 3) 3 (2 x^{2} - 6 x + 7) d x

\frac{3}{16 3 2} (4 x^{2} - 12 x + 14)^{\frac{4}{3}} + C

Schritte zur Lösung

\int (2 x - 3) 3 2 x^{2} - 6 x + 7 d x

Wende U-Substitution an

= \int \frac{u 3 u ^{2} + 5}{2 3 2} d u

Entferne die Konstante:

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

= \frac{1}{2 3 2} \cdot \int u 3 u^{2} + 5 d u

Wende U-Substitution an

= \frac{1}{2 3 2} \cdot \int \frac{3 v}{2} d v

Entferne die Konstante:

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

= \frac{1}{2 3 2} \cdot \frac{1}{2} \cdot \int 3 v d v

Wende die Potenzregel an

= \frac{1}{2 3 2} \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

Ersetze zurück

= \frac{1}{2 3 2} \cdot \frac{1}{2} \cdot \frac{3}{4} ((2 x - 3)^{2} + 5)^{\frac{4}{3}}

Vereinfache

\frac{1}{2 3 2} \cdot \frac{1}{2} \cdot \frac{3}{4} ((2 x - 3)^{2} + 5)^{\frac{4}{3}} : \frac{3}{16 3 2} (4 x^{2} - 12 x + 14)^{\frac{4}{3}}

= \frac{3}{16 3 2} (4 x^{2} - 12 x + 14)^{\frac{4}{3}}

Füge eine Konstante zur Lösung hinzu

= \frac{3}{16 3 2} (4 x^{2} - 12 x + 14)^{\frac{4}{3}} + C

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