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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

1 + x + x^{2} : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Wende U-Substitution an

= \int \frac{2}{4 u ^{2} + 3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{4 u ^{2} + 3} d u

Trigonometrische Substitution anwenden

= 2 \cdot \int \frac{sec ( v )}{2} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} (x + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x + \frac{1}{2})))

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

f^{^{'}} (x) = 5 (1 + e^{x})^{4} e^{x}

d er i v a t i v e h (w) = \frac{5 w ^{3} - w}{w}

\frac{d}{d x} ((x^{2} - 3)^{4})

\frac{d}{d x} ((a + \frac{b}{x ^{2}})^{3})

roo t s x y^{3}