integral of (\sqrt[3]{h+1}+h-1)/h

Lösung

\int \frac{3 h + 1 + h - 1}{h} d h

3 3 h + 1 + ln 3 h + 1 - 1 - \frac{1}{2} ln 4 (h + 1)^{\frac{2}{3}} + 4 3 h + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 h + 1 + 1)) + h - ln ∣ h ∣ + C

Schritte zur Lösung

\int \frac{3 h + 1 + h - 1}{h} d h

Multipliziere aus

\frac{3 h + 1 + h - 1}{h} : \frac{3 h + 1}{h} + 1 - \frac{1}{h}

= \int \frac{3 h + 1}{h} + 1 - \frac{1}{h} d h

Wende die Summenregel an:

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

= \int \frac{3 h + 1}{h} d h + \int 1 d h - \int \frac{1}{h} d h

\int \frac{3 h + 1}{h} d h = 3 3 h + 1 + ln 3 h + 1 - 1 - \frac{1}{2} ln 4 (h + 1)^{\frac{2}{3}} + 4 3 h + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 h + 1 + 1))

\int 1 d h = h

\int \frac{1}{h} d h = ln ∣ h ∣

= 3 3 h + 1 + ln 3 h + 1 - 1 - \frac{1}{2} ln 4 (h + 1)^{\frac{2}{3}} + 4 3 h + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 h + 1 + 1)) + h - ln ∣ h ∣

Füge eine Konstante zur Lösung hinzu

= 3 3 h + 1 + ln 3 h + 1 - 1 - \frac{1}{2} ln 4 (h + 1)^{\frac{2}{3}} + 4 3 h + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 h + 1 + 1)) + h - ln ∣ h ∣ + C

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