integral of 1/(sqrt(x^2-2x-8))

Lösung

\int \frac{1}{x ^{2} - 2 x - 8} d x

ln (\frac{1}{3} x^{2} - 2 x - 8 + x - 1) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 2 x - 8} d x

Vervollständige das Quadrat

x^{2} - 2 x - 8 : (x - 1)^{2} - 9

= \int \frac{1}{( x - 1 ) ^{2} - 9} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 9} d u

Trigonometrische Substitution anwenden

= \int sec (v) d v

Nutze das gemeinsame Integral :

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

= ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= ln tan (arcsec (\frac{1}{3} (x - 1))) + sec (arcsec (\frac{1}{3} (x - 1)))

Vereinfache

ln tan (arcsec (\frac{1}{3} (x - 1))) + sec (arcsec (\frac{1}{3} (x - 1))) : ln (\frac{1}{3} x^{2} - 2 x - 8 + x - 1)

= ln (\frac{1}{3} x^{2} - 2 x - 8 + x - 1)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{3} x^{2} - 2 x - 8 + x - 1) + C

\frac{\partial}{\partial x} (3 ln (x^{2} + y^{2}))

\int \frac{x ^{3}}{x - 1} d x

(8 - 6 y + e^{- 3 x}) d x - 2 d y = 0

\frac{d}{d x} ((3 x - x^{3} + 1)^{4})

\int (40 - \frac{2}{sec ( θ )}) d θ