integral of (6x+5)/(\sqrt[3]{3x^2+5x)}

Lösung

\int \frac{6 x + 5}{3 3 x ^{2} + 5 x} d x

\frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}} + C

Schritte zur Lösung

\int \frac{6 x + 5}{3 3 x ^{2} + 5 x} d x

Wende U-Substitution an

= \int \frac{u}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}} 3 u ^{2} - 25} d u

Entferne die Konstante:

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

= \frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \int \frac{u}{3 u ^{2} - 25} d u

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

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

Ersetze zurück

= \frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \frac{1}{2} \cdot \frac{3}{2} ((6 x + 5)^{2} - 25)^{\frac{2}{3}}

Vereinfache

\frac{1}{2 ^{\frac{1}{3}} \cdot 3 ^{\frac{2}{3}}} \cdot \frac{1}{2} \cdot \frac{3}{2} ((6 x + 5)^{2} - 25)^{\frac{2}{3}} : \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}}

= \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}}

Füge eine Konstante zur Lösung hinzu

= \frac{3 3}{4 3 2} (36 x^{2} + 60 x)^{\frac{2}{3}} + C

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

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t^{2} \cdot \frac{d y}{d t} + y = 0