integral of sqrt(x^2-25)

Lösung

\int x^{2} - 25 d x

25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5})) + C

Schritte zur Lösung

\int x^{2} - 25 d x

Trigonometrische Substitution anwenden

= \int 25 tan^{2} (u) sec (u) d u

Entferne die Konstante:

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

= 25 \cdot \int tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 25 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

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

= 25 \cdot \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

= 25 (- \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) ∣

= 25 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arcsec (\frac{1}{5} x)

ein

= 25 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)))

Vereinfache

25 (- ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x)) + \frac{1}{2} sec (arcsec (\frac{1}{5} x)) tan (arcsec (\frac{1}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} x)) + sec (arcsec (\frac{1}{5} x))) : 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5}))

= 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5}))

Füge eine Konstante zur Lösung hinzu

= 25 (\frac{1}{50} x x^{2} - 25 - \frac{1}{2} ln (\frac{x + x ^{2} - 25}{5})) + C

d er i v a t i v e f (x) = (x + \frac{1}{3 x})^{2}

x \to \infty lim (e^{- 6 x} cos (x))

x y^{2} y^{^{'}} = y^{3} - 3 x^{3}, y (1) = 6

a re a x = 8 y^{2}, x = 2 + 6 y^{2}

x \to - \infty lim (1 0^{x})