integral of sqrt(9+x^2)

Lösung

\int 9 + x^{2} d x

9 (\frac{1}{18} x 9 + x^{2} + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3})) + C

Schritte zur Lösung

\int 9 + x^{2} d x

Trigonometrische Substitution anwenden

= \int 9 sec^{3} (u) d u

Entferne die Konstante:

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

= 9 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{1}{3} x)

ein

= 9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} x ) ) sin ( arctan ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)))

Vereinfache

9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} x ) ) sin ( arctan ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))) : 9 (\frac{1}{18} x 9 + x^{2} + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3}))

= 9 (\frac{1}{18} x 9 + x^{2} + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3}))

Füge eine Konstante zur Lösung hinzu

= 9 (\frac{1}{18} x 9 + x^{2} + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3})) + C

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