integral of 1/(P(a-bP))

Lösung

\int \frac{1}{P ( a - b P )} d P

\frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1) + C

Schritte zur Lösung

\int \frac{1}{P ( a - b P )} d P

Wende U-Substitution an

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

u - a = a (\frac{u}{a} - 1)

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int \frac{1}{u ( \frac{u}{a} - 1 )} d u

Wende U-Substitution an

= \frac{1}{a} \cdot \int \frac{1}{v ( v - 1 )} d v

Ermittle den Partialbruch von

\frac{1}{v ( v - 1 )} : - \frac{1}{v} + \frac{1}{v - 1}

= \frac{1}{a} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} 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

= \frac{1}{a} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{a} (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Ersetze zurück

= \frac{1}{a} (- ln \frac{a - b P}{a} + ln \frac{a - b P}{a} - 1)

Wende Absoluteregel an:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{a - b P}{a} = \frac{∣ a - b P ∣}{∣ a ∣}

= \frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1) + C

\frac{\partial}{\partial y} (\frac{x ^{2}}{y + 1})

\frac{\partial}{\partial x} (- sin (2 x))

\frac{d}{d x} (\frac{e ^{3 x}}{e ^{3 x} + 2})

\frac{d}{d x} (\frac{- 3}{x ^{2}})

\int (\frac{1}{x ^{3}}) d x