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Beliebt Rechnen >

integral of 1/(f(f-u))

Lösung

\int \frac{1}{f ( f - u )} df

Lösung

\frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1) + C

Schritte zur Lösung

\int \frac{1}{f ( f - u )} df

- u + f = u (- 1 + \frac{f}{u})

= \int \frac{1}{u ( - 1 + \frac{f}{u} ) f} df

Entferne die Konstante:

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

= \frac{1}{u} \cdot \int \frac{1}{( - 1 + \frac{f}{u} ) f} df

Wende U-Substitution an

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

Setze in

v = \frac{f}{u}

ein

= \frac{1}{u} (- ln \frac{f}{u} + ln \frac{f}{u} - 1)

Wende Absoluteregel an:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{f}{u} = \frac{∣ f ∣}{∣ u ∣}

= \frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1) + C

Graph

Beliebte Beispiele

integral from 1 to 5 of (4x-4/x)

\int_{1}^{5} (4 x - \frac{4}{x}) d x

inverselaplace 2/(s^2+2s+3)

in v erse l a pl a ce \frac{2}{s ^{2} + 2 s + 3}

integral of (x^3)/((1+5x^2)^3)

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

integral of 4x^2sqrt(3x^3+2)

\int 4 x^{2} 3 x^{3} + 2 d x

derivative of (15x^6/(3x))

\frac{d}{d x} (\frac{15 x ^{6}}{3 x})