integral of (x^3)/(\sqrt[3]{x^2+4)}

Lösung

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

\frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}}) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u ^{\frac{1}{3}}}{3 3 u ^{\frac{2}{3}} + 4} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{3}}}{3 u ^{\frac{2}{3}} + 4} d u

Wende U-Substitution an

= \frac{1}{3} \cdot \int \frac{3 ( v - 4 )}{2 3 v} d v

Entferne die Konstante:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int \frac{v - 4}{3 v} d v

Multipliziere aus

\frac{v - 4}{3 v} : v^{\frac{2}{3}} - \frac{4}{3 v}

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

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} \cdot \frac{3}{2} (\int v^{\frac{2}{3}} d v - \int \frac{4}{3 v} d v)

\int v^{\frac{2}{3}} d v = \frac{3}{5} v^{\frac{5}{3}}

\int \frac{4}{3 v} d v = 6 v^{\frac{2}{3}}

= \frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} v^{\frac{5}{3}} - 6 v^{\frac{2}{3}})

Ersetze zurück

= \frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{5}{3}} - 6 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{2}{3}})

Vereinfache

\frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{5}{3}} - 6 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{2}{3}}) : \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}})

= \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}}) + C

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