integral of 1/(sqrt(x^2+4x+10))

Lösung

\int \frac{1}{x ^{2} + 4 x + 10} d x

ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 4 x + 10} d x

Vervollständige das Quadrat

x^{2} + 4 x + 10 : (x + 2)^{2} + 6

= \int \frac{1}{( x + 2 ) ^{2} + 6} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} + 6} 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}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2)))

Vereinfache

ln tan (arctan (\frac{1}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2))) : ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10)

= ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10)

Füge eine Konstante zur Lösung hinzu

= ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10) + C

y cos^{2} (x) \frac{d y}{d x} = 4 + tan^{2} (x)

\int x + 7 d x

d er i v a t i v e ln (\frac{( 4 - x )}{4 + x})

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

y^{^{'}} = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}, y (0) = 2