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Beliebt Rechnen >

integral from 1 to x of sqrt(t^2-1)

Lösung

\int_{1}^{x} t^{2} - 1 d t

Lösung

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

Schritte zur Lösung

\int_{1}^{x} t^{2} - 1 d t

Trigonometrische Substitution anwenden

= \int_{0}^{arcsec (x)} tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{arcsec (x)} (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

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

= \int_{0}^{arcsec (x)} - 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_{0}^{arcsec (x)} sec (u) d u + \int_{0}^{arcsec (x)} sec^{3} (u) d u

\int_{0}^{arcsec (x)} sec (u) d u = ln x + x^{2} - 1

\int_{0}^{arcsec (x)} sec^{3} (u) d u = \frac{x x ^{2} - 1 + ln x + x ^{2} - 1}{2}

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

Beliebte Beispiele

integral from 0 to 5 of x(5-x)

\int_{0}^{5} x (5 - x) d x

integral from infinity to 0 of xe^{-4x}

\int_{\infty}^{0} x e^{- 4 x} d x

integral from 2 to 4 of 2+x

\int_{2}^{4} 2 + x d x

integral from 0 to 4 of 12

\int_{0}^{4} 12 d x

integral from 0 to x of t^k

\int_{0}^{x} t^{k} d t