integral of 1/(sqrt(x^2-8x-9))

Lösung

\int \frac{1}{x ^{2} - 8 x - 9} d x

ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 8 x - 9} d x

Vervollständige das Quadrat

x^{2} - 8 x - 9 : (x - 4)^{2} - 25

= \int \frac{1}{( x - 4 ) ^{2} - 25} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 25} 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}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4)))

Vereinfache

ln tan (arcsec (\frac{1}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4))) : ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

= ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

\int 3 x^{\frac{1}{2}} + 3 d x

\int \frac{x ^{6}}{( 3 + x ^{7} ) ^{2}} d x

\int \frac{12 x ^{2} + 4 x - 16}{6 x ^{2} - 13 x - 5} d x

\int x^{3} + 2 x (3 x^{2} + 2) d x

\int - 3 x^{2} + 2 x d x