integral of 1/(p^2+2p)

Lösung

\int \frac{1}{p ^{2} + 2 p} d p

- \frac{1}{2} ln ∣ p + 2 ∣ + \frac{1}{2} ln ∣ p ∣ + C

Schritte zur Lösung

\int \frac{1}{p ^{2} + 2 p} d p

Vervollständige das Quadrat

p^{2} + 2 p : (p + 1)^{2} - 1

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

Wende U-Substitution an

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

Faktorisiere

u^{2} - 1 : - (- u^{2} + 1)

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = p + 1

ein

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

Vereinfache

- (\frac{ln ∣ p + 1 + 1 ∣}{2} - \frac{ln ∣ p + 1 - 1 ∣}{2}) : - \frac{1}{2} ln ∣ p + 2 ∣ + \frac{1}{2} ln ∣ p ∣

= - \frac{1}{2} ln ∣ p + 2 ∣ + \frac{1}{2} ln ∣ p ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln ∣ p + 2 ∣ + \frac{1}{2} ln ∣ p ∣ + C

a re a x = y^{2}, x = y + 6

\frac{d}{d x} (csc (x) cot (x))

d er i v a t i v e x^{2}

a re a f (x) = - x^{2} + 2, g (x) = - x - 4

\int v^{- \frac{4}{5}} d v