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integral of (4x^5)/(sqrt(4+x^2))

Lösung

\int \frac{4 x ^{5}}{4 + x ^{2}} d x

128 (\frac{1}{2} 4 + x^{2} - \frac{1}{12} (4 + x^{2})^{\frac{3}{2}} + \frac{1}{160} (4 + x^{2})^{\frac{5}{2}}) + C

Schritte zur Lösung

\int \frac{4 x ^{5}}{4 + x ^{2}} d x

Entferne die Konstante:

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

= 4 \cdot \int \frac{x ^{5}}{4 + x ^{2}} d x

Trigonometrische Substitution anwenden

= 4 \cdot \int 32 tan^{5} (u) sec (u) d u

Entferne die Konstante:

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

= 4 \cdot 32 \cdot \int tan^{5} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \cdot 32 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

Wende U-Substitution an

= 4 \cdot 32 \cdot \int (- 1 + v^{2})^{2} d v

Multipliziere aus

(- 1 + v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= 4 \cdot 32 \cdot \int 1 - 2 v^{2} + v^{4} 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

= 4 \cdot 32 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= 4 \cdot 32 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Ersetze zurück

= 4 \cdot 32 (sec (arctan (\frac{1}{2} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} x ) )}{5})

Vereinfache

4 \cdot 32 (sec (arctan (\frac{1}{2} x)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} x ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} x ) )}{5}) : 128 (\frac{1}{2} 4 + x^{2} - \frac{1}{12} (4 + x^{2})^{\frac{3}{2}} + \frac{1}{160} (4 + x^{2})^{\frac{5}{2}})

= 128 (\frac{1}{2} 4 + x^{2} - \frac{1}{12} (4 + x^{2})^{\frac{3}{2}} + \frac{1}{160} (4 + x^{2})^{\frac{5}{2}})

Füge eine Konstante zur Lösung hinzu

= 128 (\frac{1}{2} 4 + x^{2} - \frac{1}{12} (4 + x^{2})^{\frac{3}{2}} + \frac{1}{160} (4 + x^{2})^{\frac{5}{2}}) + C

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