integral 1/((4-x^2))

Lösung

integral

\frac{1}{( 4 - x ^{2} )}

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

Schritte zur Lösung

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

Wende integrale Substitution an

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

Nutze das gemeinsame Integral :

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

ein

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

Vereinfache

\frac{1}{2} (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}) : \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{1}{( x ^{2} + x + 1 )}

in t e g r a l cos^{3} (x) - cos (x)

in t e g r a l sin (2 - 4 x)

in t e g r a l \frac{x}{( x ^{2} + a ^{2} )}

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