integral of (sqrt(x^2-1))

Lösung

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

\frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1 + C

Schritte zur Lösung

\int x^{2} - 1 d x

Trigonometrische Substitution anwenden

= \int tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

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

Setze in

u = arcsec (x)

ein

= - ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ + \frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣

Vereinfache

- ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ + \frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣ : \frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1

= \frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1 + C

\int 2 t \cdot e^{t} d t

(1 + e^{x}) y \frac{d y}{d x} = e^{y}

d er i v a t i v e \frac{sec ( 9 x )}{1 + sec ( 9 x )}

\frac{d}{d x} (3 sin (x) - 4 sin^{3} (x))

n = 1 \sum \infty \frac{3}{2 n}