integral of sqrt(6+7t^2)

Lösung

\int 6 + 7 t^{2} d t

\frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6} + C

Schritte zur Lösung

\int 6 + 7 t^{2} d t

Trigonometrische Substitution anwenden

= \int \frac{6}{7} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{6}{7} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{6}{7} (\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{6}{7} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{7}{6} t)

ein

= \frac{6}{7} \frac{sec ^{2} ( arctan ( \frac{7}{6} t ) ) sin ( arctan ( \frac{7}{6} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{6} t)) + sec (arctan (\frac{7}{6} t))

Vereinfache

\frac{6}{7} \frac{sec ^{2} ( arctan ( \frac{7}{6} t ) ) sin ( arctan ( \frac{7}{6} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{6} t)) + sec (arctan (\frac{7}{6} t)) : \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6}

= \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} t 7 t^{2} + 6 + \frac{3}{7} ln \frac{7}{6} t + \frac{7 t ^{2} + 6}{6} + C