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

Lösung

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

ln (\frac{1}{4} x^{2} - 8 x + x - 4) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} - 8 x : (x - 4)^{2} - 16

= \int \frac{1}{( x - 4 ) ^{2} - 16} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 16} 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}{4} (x - 4))) + sec (arcsec (\frac{1}{4} (x - 4)))

Vereinfache

ln tan (arcsec (\frac{1}{4} (x - 4))) + sec (arcsec (\frac{1}{4} (x - 4))) : ln (\frac{1}{4} x^{2} - 8 x + x - 4)

= ln (\frac{1}{4} x^{2} - 8 x + x - 4)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{4} x^{2} - 8 x + x - 4) + C

\frac{\partial}{\partial x} (2 y cos (x y))

x \to - 3 lim (- 4 x + 8)

\frac{\partial}{\partial x} (\frac{1}{ln ( x )})

\frac{\partial}{\partial x} (e^{x^{2} + y^{2} + z^{2}})

\frac{\partial}{\partial x} (\frac{x}{x + y})