integral of 1/(sqrt(x^2-16x))

Lösung

\int \frac{1}{x ^{2} - 16 x} d x

ln (\frac{1}{8} x^{2} - 16 x + x - 8) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 16 x} d x

Vervollständige das Quadrat

x^{2} - 16 x : (x - 8)^{2} - 64

= \int \frac{1}{( x - 8 ) ^{2} - 64} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 64} 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}{8} (x - 8))) + sec (arcsec (\frac{1}{8} (x - 8)))

Vereinfache

ln tan (arcsec (\frac{1}{8} (x - 8))) + sec (arcsec (\frac{1}{8} (x - 8))) : ln (\frac{1}{8} x^{2} - 16 x + x - 8)

= ln (\frac{1}{8} x^{2} - 16 x + x - 8)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{8} x^{2} - 16 x + x - 8) + C

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

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\int (\frac{6 + x ^{2}}{x}) d x