integral of 1/(3-6x-x^2)

Lösung

\int \frac{1}{3 - 6 x - x ^{2}} d x

\frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1) + C

Schritte zur Lösung

\int \frac{1}{3 - 6 x - x ^{2}} d x

Vervollständige das Quadrat

3 - 6 x - x^{2} : - (x + 3)^{2} + 12

= \int \frac{1}{- ( x + 3 ) ^{2} + 12} d x

Wende U-Substitution an

= \int \frac{1}{- u ^{2} + 12} d u

Wende integrale Substitution an

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

Ersetze zurück

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

Vereinfache

\frac{1}{2 3} \frac{ln \frac{x + 3}{2 3} + 1}{2} - \frac{ln \frac{x + 3}{2 3} - 1}{2} : \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1)

= \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1) + C

x \to 0 lim (5 x + 4)

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 7))

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

l a pl a ce t r an s f or m z

\int \frac{12}{3 x + x x} d x