integral of 1/(g-k/m x^2)

Lösung

\int \frac{1}{g - \frac{k}{m} x ^{2}} d x

\frac{m}{2 g k} (ln \frac{k}{g m} x + 1 - ln \frac{k}{g m} x - 1) + C

Schritte zur Lösung

\int \frac{1}{g - \frac{k}{m} x ^{2}} d x

Faktorisiere

\frac{1}{g - \frac{k}{m} x ^{2}}

= \int \frac{1}{g - \frac{k x ^{2}}{m}} d x

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Setze in

u = \frac{\frac{k}{m}}{g} x

ein

= \frac{m}{g k} \frac{ln \frac{\frac{k}{m}}{g} x + 1}{2} - \frac{ln \frac{\frac{k}{m}}{g} x - 1}{2}

Vereinfache

\frac{m}{g k} \frac{ln \frac{\frac{k}{m}}{g} x + 1}{2} - \frac{ln \frac{\frac{k}{m}}{g} x - 1}{2} : \frac{m}{2 g k} (ln \frac{k}{g m} x + 1 - ln \frac{k}{g m} x - 1)

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

Füge eine Konstante zur Lösung hinzu

= \frac{m}{2 g k} (ln \frac{k}{g m} x + 1 - ln \frac{k}{g m} x - 1) + C

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