integral of sqrt(12+4x^2)

Lösung

\int 12 + 4 x^{2} d x

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

Schritte zur Lösung

\int 12 + 4 x^{2} d x

Trigonometrische Substitution anwenden

= \int 6 sec^{3} (u) d u

Entferne die Konstante:

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

= 6 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 6 (\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) ∣

= 6 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{2}{2 3} x)

ein

= 6 \frac{sec ^{2} ( arctan ( \frac{2}{2 3} x ) ) sin ( arctan ( \frac{2}{2 3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2 3} x)) + sec (arctan (\frac{2}{2 3} x))

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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