integral of 1/((5-4x-x^2))

Lösung

\int \frac{1}{( 5 - 4 x - x ^{2} )} d x

\frac{1}{6} (ln \frac{x + 2}{3} + 1 - ln \frac{x + 2}{3} - 1) + C

Schritte zur Lösung

\int \frac{1}{5 - 4 x - x ^{2}} d x

Vervollständige das Quadrat

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

= \int \frac{1}{- ( x + 2 ) ^{2} + 9} d x

Wende U-Substitution an

= \int \frac{1}{- u ^{2} + 9} d u

Wende integrale Substitution an

= \int \frac{1}{3 ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{1}{- v ^{2} + 1} d v

Nutze das gemeinsame Integral :

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Ersetze zurück

= \frac{1}{3} (\frac{ln \frac{x + 2}{3} + 1}{2} - \frac{ln \frac{x + 2}{3} - 1}{2})

Vereinfache

\frac{1}{3} (\frac{ln \frac{x + 2}{3} + 1}{2} - \frac{ln \frac{x + 2}{3} - 1}{2}) : \frac{1}{6} (ln \frac{x + 2}{3} + 1 - ln \frac{x + 2}{3} - 1)

= \frac{1}{6} (ln \frac{x + 2}{3} + 1 - ln \frac{x + 2}{3} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} (ln \frac{x + 2}{3} + 1 - ln \frac{x + 2}{3} - 1) + C

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