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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= 2 \cdot \int \frac{cos ( u )}{sin ( u ) + cos ( u )} d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 2 \cdot \int \frac{1}{2} + \frac{cos ( u ) - sin ( u )}{2 ( sin ( u ) + cos ( u ) )} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{2} d u + \int \frac{cos ( u ) - sin ( u )}{2 ( sin ( u ) + cos ( u ) )} d u)

\int \frac{1}{2} d u = \frac{1}{2} u

\int \frac{cos ( u ) - sin ( u )}{2 ( sin ( u ) + cos ( u ) )} d u = \frac{1}{2} ln ∣ sin (u) + cos (u) ∣

= 2 (\frac{1}{2} u + \frac{1}{2} ln ∣ sin (u) + cos (u) ∣)

Setze in

u = arcsin (x)

ein

= 2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln ∣ sin (arcsin (x)) + cos (arcsin (x)) ∣)

Vereinfache

2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln ∣ sin (arcsin (x)) + cos (arcsin (x)) ∣) : 2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln x + 1 - x^{2})

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

Füge eine Konstante zur Lösung hinzu

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

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

\int 243 x - 482 d x

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