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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{tan ^{5} ( ln ( x ) )}{x} d x

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