integral of 1/(64-x^2)

Lösung

\int \frac{1}{64 - x ^{2}} d x

\frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1) + C

Schritte zur Lösung

\int \frac{1}{64 - x ^{2}} d x

Wende integrale Substitution an

= \int \frac{1}{8 ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{8} \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}{8} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Setze in

u = \frac{x}{8}

ein

= \frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2})

Vereinfache

\frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2}) : \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1)

= \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1) + C

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