integral of 1/(sqrt(x^2-2x+10))

Lösung

\int \frac{1}{x ^{2} - 2 x + 10} d x

ln (\frac{1}{3} x - 1 + x^{2} - 2 x + 10) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 2 x + 10} d x

Vervollständige das Quadrat

x^{2} - 2 x + 10 : (x - 1)^{2} + 9

= \int \frac{1}{( x - 1 ) ^{2} + 9} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} + 9} 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 (arctan (\frac{1}{3} (x - 1))) + sec (arctan (\frac{1}{3} (x - 1)))

Vereinfache

ln tan (arctan (\frac{1}{3} (x - 1))) + sec (arctan (\frac{1}{3} (x - 1))) : ln (\frac{1}{3} x - 1 + x^{2} - 2 x + 10)

= ln (\frac{1}{3} x - 1 + x^{2} - 2 x + 10)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{3} x - 1 + x^{2} - 2 x + 10) + C

49 y^{^{''}} + 28 y^{^{'}} + 4 y = 0

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 94

x \to 0 lim (x \cdot sin (\frac{π}{x}))

m = \frac{9}{5 + h} - \frac{9}{5}

n = 1 \sum \infty e^{- n} + 1