integral of sqrt(t^2+1)

Lösung

\int t^{2} + 1 d t

\frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2} + C

Schritte zur Lösung

\int t^{2} + 1 d t

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

Setze in

u = arctan (t)

ein

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

Vereinfache

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

= \frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2}

Füge eine Konstante zur Lösung hinzu

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

\frac{d x}{d t} = x - x^{3}

\int_{0}^{\infty} e^{- 2 a x} d x

x \to 9 + lim (\frac{x + 8}{x + 9})

t an g e n t f (x) = \frac{x - 7}{2 x - 3}, at x = 1

\int \frac{arccos ( x ) - x}{1 - x ^{2}} d x