integral of 1/(sqrt(x^2-4x+13))

Lösung

\int \frac{1}{x ^{2} - 4 x + 13} d x

ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 4 x + 13} d x

Vervollständige das Quadrat

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

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

Vereinfache

ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2))) : ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

= ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

\frac{\partial}{\partial x} (4 x^{8 y})

\int x 2 - 1 - x^{2} d x

\int (arccos (x))^{2} d x

d er i v a t i v e y = x^{2} 16 - x^{2}

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