integral of 1/(sqrt(36x^2+2))

Lösung

\int \frac{1}{36 x ^{2} + 2} d x

\frac{1}{6} ln 32 x + 1 + 18 x^{2} + C

Schritte zur Lösung

\int \frac{1}{36 x ^{2} + 2} d x

Trigonometrische Substitution anwenden

= \int \frac{sec ( u )}{6} d u

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = arctan (\frac{6}{2} x)

ein

= \frac{1}{6} ln tan (arctan (\frac{6}{2} x)) + sec (arctan (\frac{6}{2} x))

Vereinfache

\frac{1}{6} ln tan (arctan (\frac{6}{2} x)) + sec (arctan (\frac{6}{2} x)) : \frac{1}{6} ln 32 x + 1 + 18 x^{2}

= \frac{1}{6} ln 32 x + 1 + 18 x^{2}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln 32 x + 1 + 18 x^{2} + C

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