integral of sqrt(u^2+a^2)

Lösung

\int u^{2} + a^{2} d u

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

Schritte zur Lösung

\int u^{2} + a^{2} d u

Trigonometrische Substitution anwenden

= \int a^{2} sec^{3} (v) d v

Entferne die Konstante:

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

= a^{2} \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

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

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

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

Setze in

v = arctan (\frac{1}{a} u)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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