integral of sqrt(9t^2+4)

Lösung

\int 9 t^{2} + 4 d t

\frac{1}{6} (3 t 9 t^{2} + 4 + 4 ln (\frac{3 t + 9 t ^{2} + 4}{2})) + C

Schritte zur Lösung

\int 9 t^{2} + 4 d t

Trigonometrische Substitution anwenden

= \int \frac{4}{3} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{4}{3} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{4}{3} (\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{4}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

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

ein

= \frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} t ) ) sin ( arctan ( \frac{3}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} t)) + sec (arctan (\frac{3}{2} t)))

Vereinfache

\frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} t ) ) sin ( arctan ( \frac{3}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} t)) + sec (arctan (\frac{3}{2} t))) : \frac{1}{6} (3 t 9 t^{2} + 4 + 4 ln (\frac{3 t + 9 t ^{2} + 4}{2}))

= \frac{1}{6} (3 t 9 t^{2} + 4 + 4 ln (\frac{3 t + 9 t ^{2} + 4}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} (3 t 9 t^{2} + 4 + 4 ln (\frac{3 t + 9 t ^{2} + 4}{2})) + C

\frac{\partial}{\partial x} (\frac{9}{x ^{2} + y ^{2}})

\int (5 x^{4} - 6 x^{2} + 3) d x

\frac{d}{d x} (ln (e^{x} + 3))

\int tan (9 x) sec^{\frac{5}{4}} (9 x) d x

t \to - 3 lim (\frac{t - 2}{t + 5})