integral of sqrt(1+(4/(3x^{1/3)))^2}

Lösung

\int 1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} d x

\frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

Schritte zur Lösung

\int 1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} d x

Vereinfache

1 + (\frac{4}{3 x ^{\frac{1}{3}}})^{2} : \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}}

= \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{( 9 x ^{\frac{2}{3}} + 16 ) ^{\frac{1}{2}}}{( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Wende U-Substitution an

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{2}}}{6} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot \int u^{\frac{1}{2}} d u

Wende die Potenzregel an

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Setze in

u = 9 x^{\frac{2}{3}} + 16

ein

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Vereinfache

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{27} (9 x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

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