integral of 1/(sqrt(3x^2+5x+1))

Lösung

\int \frac{1}{3 x ^{2} + 5 x + 1} d x

\frac{1}{3} ln \frac{36 x ^{2} + 60 x + 12}{13} + \frac{6}{13} x + \frac{5}{13} + C

Schritte zur Lösung

\int \frac{1}{3 x ^{2} + 5 x + 1} d x

Vervollständige das Quadrat

3 x^{2} + 5 x + 1 : 3 (x + \frac{5}{6})^{2} - \frac{13}{12}

= \int \frac{1}{3 ( x + \frac{5}{6} ) ^{2} - \frac{13}{12}} d x

Wende U-Substitution an

= \int \frac{2 3}{36 u ^{2} - 13} d u

Entferne die Konstante:

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

= 23 \cdot \int \frac{1}{36 u ^{2} - 13} d u

Trigonometrische Substitution anwenden

= 23 \cdot \int \frac{sec ( v )}{6} d v

Entferne die Konstante:

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

= 23 \frac{1}{6} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= 23 \frac{1}{6} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= 23 \frac{1}{6} ln tan (arcsec (\frac{6}{13} (x + \frac{5}{6}))) + sec (arcsec (\frac{6}{13} (x + \frac{5}{6})))

Vereinfache

23 \frac{1}{6} ln tan (arcsec (\frac{6}{13} (x + \frac{5}{6}))) + sec (arcsec (\frac{6}{13} (x + \frac{5}{6}))) : \frac{1}{3} ln \frac{36 x ^{2} + 60 x + 12}{13} + \frac{6}{13} x + \frac{5}{13}

= \frac{1}{3} ln \frac{36 x ^{2} + 60 x + 12}{13} + \frac{6}{13} x + \frac{5}{13}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} ln \frac{36 x ^{2} + 60 x + 12}{13} + \frac{6}{13} x + \frac{5}{13} + C

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