integral of 1/(sqrt(a^2+b^2x^2))

Lösung

\int \frac{1}{a ^{2} + b ^{2} x ^{2}} d x

\frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣}) + C

Schritte zur Lösung

\int \frac{1}{a ^{2} + b ^{2} x ^{2}} d x

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = arctan (\frac{b}{a} x)

ein

= \frac{1}{b} ln tan (arctan (\frac{b}{a} x)) + sec (arctan (\frac{b}{a} x))

Vereinfache

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

= \frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣}) + C

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