integral of 1/(sqrt(x^2-4x-3))

Lösung

\int \frac{1}{x ^{2} - 4 x - 3} d x

ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 4 x - 3} d x

Vervollständige das Quadrat

x^{2} - 4 x - 3 : (x - 2)^{2} - 7

= \int \frac{1}{( x - 2 ) ^{2} - 7} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 7} 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}{7} (x - 2))) + sec (arcsec (\frac{1}{7} (x - 2)))

Vereinfache

ln tan (arcsec (\frac{1}{7} (x - 2))) + sec (arcsec (\frac{1}{7} (x - 2))) : ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2)

= ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2)

Füge eine Konstante zur Lösung hinzu

= ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2) + C

\int \frac{2 x ^{2}}{( x ^{2} + 5 ) ^{2}} d x

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

in t e g r a l cos^{2} (t)

\int \frac{1}{2}

\int sin (a x) e^{b x + c} d x