integral of 1/(1+\sqrt[3]{x+2)}

Lösung

\int \frac{1}{1 + 3 x + 2} d x

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

Schritte zur Lösung

\int \frac{1}{1 + 3 x + 2} d x

Wende U-Substitution an

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 3 \cdot \int \frac{( w - 1 ) ^{2}}{w} d w

Multipliziere aus

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

= 3 \cdot \int w - 2 + \frac{1}{w} d w

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int w d w - \int 2 d w + \int \frac{1}{w} d w)

\int w d w = \frac{w ^{2}}{2}

\int 2 d w = 2 w

\int \frac{1}{w} d w = ln ∣ w ∣

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

Ersetze zurück

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

Füge eine Konstante zur Lösung hinzu

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

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