integral of 1/(sqrt(x^2-4x+8))

Lösung

\int \frac{1}{x ^{2} - 4 x + 8} d x

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

Schritte zur Lösung

\int \frac{1}{x ^{2} - 4 x + 8} d x

Vervollständige das Quadrat

x^{2} - 4 x + 8 : (x - 2)^{2} + 4

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

Wende U-Substitution an

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (2 x^{x})

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

\int - 4 sin (5 x) sin (x) d x

\frac{d}{d x} (\frac{1}{1 + sin ( x )})

\int \frac{3}{4} t + 6 d t