integral of 1/(sqrt(x^2+4x+29))

Lösung

\int \frac{1}{x ^{2} + 4 x + 29} d x

ln (\frac{1}{5} x + 2 + x^{2} + 4 x + 29) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 4 x + 29} d x

Vervollständige das Quadrat

x^{2} + 4 x + 29 : (x + 2)^{2} + 25

= \int \frac{1}{( x + 2 ) ^{2} + 25} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} + 25} 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}{5} (x + 2))) + sec (arctan (\frac{1}{5} (x + 2)))

Vereinfache

ln tan (arctan (\frac{1}{5} (x + 2))) + sec (arctan (\frac{1}{5} (x + 2))) : ln (\frac{1}{5} x + 2 + x^{2} + 4 x + 29)

= ln (\frac{1}{5} x + 2 + x^{2} + 4 x + 29)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{5} x + 2 + x^{2} + 4 x + 29) + C

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