integral 1/((2-y^2))

Lösung

integral

\frac{1}{( 2 - y ^{2} )}

\frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1) + C

Schritte zur Lösung

\int \frac{1}{2 - y ^{2}} d y

Wende integrale Substitution an

= \int \frac{1}{2 ( 1 - u ^{2} )} d u

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Setze in

u = \frac{y}{2}

ein

= \frac{1}{2} \frac{ln \frac{y}{2} + 1}{2} - \frac{ln \frac{y}{2} - 1}{2}

Vereinfache

\frac{1}{2} \frac{ln \frac{y}{2} + 1}{2} - \frac{ln \frac{y}{2} - 1}{2} : \frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

= \frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1) + C

in t e g r a l \frac{1}{( 1 + sin ( x ) )}

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in t e g r a l \frac{( x + 3 )}{( x - 1 )}