integral of 1/(sqrt(4x^2+12x+16))

Lösung

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

\frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7} + C

Schritte zur Lösung

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

Vervollständige das Quadrat

4 x^{2} + 12 x + 16 : 4 (x + \frac{3}{2})^{2} + 7

= \int \frac{1}{4 ( x + \frac{3}{2} ) ^{2} + 7} d x

Wende U-Substitution an

= \int \frac{1}{4 u ^{2} + 7} d u

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

= \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 (arctan (\frac{2}{7} (x + \frac{3}{2}))) + sec (arctan (\frac{2}{7} (x + \frac{3}{2})))

Vereinfache

\frac{1}{2} ln tan (arctan (\frac{2}{7} (x + \frac{3}{2}))) + sec (arctan (\frac{2}{7} (x + \frac{3}{2}))) : \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7}

= \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{2}{7} x + \frac{3}{7} + \frac{4 x ^{2} + 12 x + 16}{7} + C

s l o p e (7, - 5), (- 7, - 5)

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