integral of u/(1+u)

Lösung

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

1 + u - ln ∣ 1 + u ∣ + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{v - 1}{v} d v

Multipliziere aus

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \int 1 - \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

= \int 1 d v - \int \frac{1}{v} d v

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= v - ln ∣ v ∣

Setze in

v = 1 + u

ein

= 1 + u - ln ∣ 1 + u ∣

Füge eine Konstante zur Lösung hinzu

= 1 + u - ln ∣ 1 + u ∣ + C

\int \frac{1}{9 - 3 x} d x

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