integral of 1/(sqrt(x^2+8x+3))

Lösung

\int \frac{1}{x ^{2} + 8 x + 3} d x

ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 8 x + 3} d x

Vervollständige das Quadrat

x^{2} + 8 x + 3 : (x + 4)^{2} - 13

= \int \frac{1}{( x + 4 ) ^{2} - 13} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 13} 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 (arcsec (\frac{1}{13} (x + 4))) + sec (arcsec (\frac{1}{13} (x + 4)))

Vereinfache

ln tan (arcsec (\frac{1}{13} (x + 4))) + sec (arcsec (\frac{1}{13} (x + 4))) : ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4)

= ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4)

Füge eine Konstante zur Lösung hinzu

= ln \frac{1}{13} (x^{2} + 8 x + 3) + \frac{1}{13} (x + 4) + C

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