integral of 1/(u(u+1))

Lösung

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

ln ∣ u ∣ - ln ∣ u + 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{u ( u + 1 )} : \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

\int \frac{1}{3 x ^{2} - 5} d x

\int \frac{8 x - 5}{3 x ^{2} - 5 x + 2} d x

\int \frac{3 a}{4 t} d t

\int \frac{1}{2 x ^{2} + 5 x + 3} d x

\int \frac{d}{1} (T) d x