integral of 1/(sqrt(4x^2+4x+17))

Lösung

\int \frac{1}{4 x ^{2} + 4 x + 17} d x

\frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4}) + C

Schritte zur Lösung

\int \frac{1}{4 x ^{2} + 4 x + 17} d x

Vervollständige das Quadrat

4 x^{2} + 4 x + 17 : 4 (x + \frac{1}{2})^{2} + 16

= \int \frac{1}{4 ( x + \frac{1}{2} ) ^{2} + 16} d x

Wende U-Substitution an

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

Vereinfache

\frac{1}{2} ln tan (arctan (\frac{1}{2} (x + \frac{1}{2}))) + sec (arctan (\frac{1}{2} (x + \frac{1}{2}))) : \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4})

= \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4}) + C

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