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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Vereinfache

ln ∣ u + 2 ∣ + \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣ : \frac{1}{2} ln ∣ 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 \frac{9 x}{1 + x ^{2}} d x

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in t e g r a l e^{5 x}

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