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Beliebt Rechnen >

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

Lösung

\int \frac{4 x}{4 + x} d x

Lösung

32 (\frac{1}{24} x^{\frac{3}{4}} - \frac{1}{2} 4 x + arctan (\frac{1}{2} 4 x)) + C

Schritte zur Lösung

\int \frac{4 x}{4 + x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende integrale Substitution an

= 4 \cdot \int \frac{8 v ^{4}}{1 + v ^{2}} d v

Entferne die Konstante:

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

= 4 \cdot 8 \cdot \int \frac{v ^{4}}{1 + v ^{2}} d v

Lange Division

\frac{v ^{4}}{1 + v ^{2}} : v^{2} - 1 + \frac{1}{v ^{2} + 1}

= 4 \cdot 8 \cdot \int v^{2} - 1 + \frac{1}{v ^{2} + 1} 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 8 (\int v^{2} d v - \int 1 d v + \int \frac{1}{v ^{2} + 1} d v)

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

\int 1 d v = v

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot 8 (\frac{v ^{3}}{3} - v + arctan (v))

Ersetze zurück

= 4 \cdot 8 \frac{( \frac{4 x}{2} ) ^{3}}{3} - \frac{4 x}{2} + arctan (\frac{4 x}{2})

Vereinfache

4 \cdot 8 \frac{( \frac{4 x}{2} ) ^{3}}{3} - \frac{4 x}{2} + arctan (\frac{4 x}{2}) : 32 (\frac{1}{24} x^{\frac{3}{4}} - \frac{1}{2} 4 x + arctan (\frac{1}{2} 4 x))

= 32 (\frac{1}{24} x^{\frac{3}{4}} - \frac{1}{2} 4 x + arctan (\frac{1}{2} 4 x))

Füge eine Konstante zur Lösung hinzu

= 32 (\frac{1}{24} x^{\frac{3}{4}} - \frac{1}{2} 4 x + arctan (\frac{1}{2} 4 x)) + C

Graph

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