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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

3 + 2 x + x^{2} : (x + 1)^{2} + 2

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= \int sec (v) d v

Nutze das gemeinsame Integral :

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1)))

Vereinfache

ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1))) : ln \frac{1}{2} (x + 1) + \frac{1}{2} (x^{2} + 2 x + 3)

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{3} cos (x) d x

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