integral of 1/(u+u^2)

Lösung

\int \frac{1}{u + u ^{2}} d u

ln ∣ u ∣ - ln ∣ u + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{u + u ^{2}} d u

Ermittle den Partialbruch von

\frac{1}{u + u ^{2}} : \frac{1}{u} - \frac{1}{u + 1}

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= ln ∣ u ∣ - ln ∣ u + 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ u ∣ - ln ∣ u + 1 ∣ + C

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