integral of 1/(7-4x-x^2)

Lösung

\int \frac{1}{7 - 4 x - x ^{2}} d x

\frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1) + C

Schritte zur Lösung

\int \frac{1}{7 - 4 x - x ^{2}} d x

Vervollständige das Quadrat

7 - 4 x - x^{2} : - (x + 2)^{2} + 11

= \int \frac{1}{- ( x + 2 ) ^{2} + 11} d x

Wende U-Substitution an

= \int \frac{1}{- u ^{2} + 11} d u

Wende integrale Substitution an

= \int \frac{1}{11 ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= \frac{1}{11} \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}{11} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Ersetze zurück

= \frac{1}{11} \frac{ln \frac{x + 2}{11} + 1}{2} - \frac{ln \frac{x + 2}{11} - 1}{2}

Vereinfache

\frac{1}{11} \frac{ln \frac{x + 2}{11} + 1}{2} - \frac{ln \frac{x + 2}{11} - 1}{2} : \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1)

= \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1) + C

in v erse l a pl a ce \frac{( s + 1 )}{( s ^{2} + 2 s + 9 )}

a re a x = 3 y, x = 2 + y^{2}

d er i v a t i v e 6 u^{- 1}

x \to 0 lim (\frac{( e ^{x} )}{x})

\frac{\partial}{\partial y} (5 - 2 x^{2} - y^{2})