integral of 1/(sqrt(x^2-2x+17))

Lösung

\int \frac{1}{x ^{2} - 2 x + 17} d x

ln (\frac{1}{4} x - 1 + x^{2} - 2 x + 17) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 2 x + 17} d x

Vervollständige das Quadrat

x^{2} - 2 x + 17 : (x - 1)^{2} + 16

= \int \frac{1}{( x - 1 ) ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} + 16} 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}{4} (x - 1))) + sec (arctan (\frac{1}{4} (x - 1)))

Vereinfache

ln tan (arctan (\frac{1}{4} (x - 1))) + sec (arctan (\frac{1}{4} (x - 1))) : ln (\frac{1}{4} x - 1 + x^{2} - 2 x + 17)

= ln (\frac{1}{4} x - 1 + x^{2} - 2 x + 17)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{4} x - 1 + x^{2} - 2 x + 17) + C

x \to 0 lim (ln (x^{2}))

d er i v a t i v e ln (3 x^{2} + 7 y^{2})

\frac{d y}{d x} = 5 y x

x \to 3 lim (tan (\frac{π x}{4}))

l a pl a ce t r an s f or m \frac{t ^{4}}{7}