integral of sqrt(1+4*x^2)

Lösung

\int 1 + 4 \cdot x^{2} d x

\frac{1}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C

Schritte zur Lösung

\int 1 + 4 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{1}{2} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{2} (\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) ∣

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

Setze in

u = arctan (2 x)

ein

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

Vereinfache

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 x ) ) sin ( arctan ( 2 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 x)) + sec (arctan (2 x)) ∣) : \frac{1}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

= \frac{1}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C

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