integral of 1/(y^{1/2)+y^{1/3}}

Lösung

\int \frac{1}{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}} d y

6 \frac{( y ^{\frac{1}{6}} + 1 ) ^{3}}{3} - \frac{3 ( y ^{\frac{1}{6}} + 1 ) ^{2}}{2} + 3 (y^{\frac{1}{6}} + 1) - ln y^{\frac{1}{6}} + 1 + C

Schritte zur Lösung

\int \frac{1}{y ^{\frac{1}{2}} + y ^{\frac{1}{3}}} d y

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 6 \cdot \int \frac{( v - 1 ) ^{3}}{v} d v

Multipliziere aus

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

= 6 \cdot \int v^{2} - 3 v + 3 - \frac{1}{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

= 6 (\int v^{2} d v - \int 3 v d v + \int 3 d v - \int \frac{1}{v} d v)

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

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

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 6 (\frac{v ^{3}}{3} - \frac{3 v ^{2}}{2} + 3 v - ln ∣ v ∣)

Ersetze zurück

= 6 \frac{( y ^{\frac{1}{6}} + 1 ) ^{3}}{3} - \frac{3 ( y ^{\frac{1}{6}} + 1 ) ^{2}}{2} + 3 (y^{\frac{1}{6}} + 1) - ln y^{\frac{1}{6}} + 1

Füge eine Konstante zur Lösung hinzu

= 6 \frac{( y ^{\frac{1}{6}} + 1 ) ^{3}}{3} - \frac{3 ( y ^{\frac{1}{6}} + 1 ) ^{2}}{2} + 3 (y^{\frac{1}{6}} + 1) - ln y^{\frac{1}{6}} + 1 + C

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