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

Lösung

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

ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

Schritte zur Lösung

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

Wende integrale Substitution an

= \int \frac{1}{u ^{2} - 6 u + 8} d u

Vervollständige das Quadrat

u^{2} - 6 u + 8 : (u - 3)^{2} - 1

= \int \frac{1}{( u - 3 ) ^{2} - 1} d u

Wende U-Substitution an

= \int \frac{1}{v ^{2} - 1} d v

Trigonometrische Substitution anwenden

= \int sec (w) d w

Nutze das gemeinsame Integral :

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

= ln ∣ tan (w) + sec (w) ∣

Ersetze zurück

= ln ∣ tan (arcsec (2 x - 3)) + sec (arcsec (2 x - 3)) ∣

Vereinfache

ln ∣ tan (arcsec (2 x - 3)) + sec (arcsec (2 x - 3)) ∣ : ln 2 x^{2} - 3 x + 2 + 2 x - 3

= ln 2 x^{2} - 3 x + 2 + 2 x - 3

Füge eine Konstante zur Lösung hinzu

= ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

(x^{2} - 1) \frac{d y}{d x} + 2 y = (x + 1)^{2}

d er i v a t i v e f (x) = (2 x + 3)^{2}

\int_{0}^{2} 2 x^{3} x^{2} + 4 d x

x \to 1 lim (x^{3} - 3)

t an g e n t f (x) = x^{2} + x, at x = - 1, x = 2