integral of sqrt(1+(6x)^2)

Lösung

\int 1 + (6 x)^{2} d x

\frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2}) + C

Schritte zur Lösung

\int 1 + (6 x)^{2} d x

Vereinfache

1 + (6 x)^{2} : 1 + 36 x^{2}

= \int 1 + 36 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{1}{6} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{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) ∣

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

Setze in

u = arctan (6 x)

ein

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

Vereinfache

\frac{1}{6} (\frac{sec ^{2} ( arctan ( 6 x ) ) sin ( arctan ( 6 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (6 x)) + sec (arctan (6 x)) ∣) : \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2})

= \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2}) + C

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