integral of 1/(u(u+2))

Lösung

\int \frac{1}{u ( u + 2 )} d u

\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣ + C

Schritte zur Lösung

\int \frac{1}{u ( u + 2 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

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

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣ + C

\int 2 x^{2} cos (5 x) d x

\int \frac{4 x ^{2} + x + 27}{x ^{3} + 9 x} d x

\frac{d y}{d x} + x (y^{2} + y) = 0

\int \frac{2 tan ( x )}{1 - tan ^{2} ( x )} d x

\frac{d}{d x} (tan (2 x + 1))