integral of 1/(sqrt(4x^2+12x+5))

Lösung

\int \frac{1}{4 x ^{2} + 12 x + 5} d x

\frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2} + C

Schritte zur Lösung

\int \frac{1}{4 x ^{2} + 12 x + 5} d x

Vervollständige das Quadrat

4 x^{2} + 12 x + 5 : 4 (x + \frac{3}{2})^{2} - 4

= \int \frac{1}{4 ( x + \frac{3}{2} ) ^{2} - 4} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{2} ln tan (arcsec (x + \frac{3}{2})) + sec (arcsec (x + \frac{3}{2}))

Vereinfache

\frac{1}{2} ln tan (arcsec (x + \frac{3}{2})) + sec (arcsec (x + \frac{3}{2})) : \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2}

= \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{4 x ^{2} + 12 x + 5}{2} + x + \frac{3}{2} + C

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