integral of 1/(tsqrt(25-t^2))

Lösung

\int \frac{1}{t 25 - t ^{2}} d t

- \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1) + C

Schritte zur Lösung

\int \frac{1}{t 25 - t ^{2}} d t

Wende U-Substitution an

= \int - \frac{1}{- u ^{2} + 25} d u

Entferne die Konstante:

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

= - \int \frac{1}{- u ^{2} + 25} d u

Wende integrale Substitution an

= - \int \frac{1}{5 ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

= - \frac{1}{5} \cdot \int \frac{1}{- v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= - \frac{1}{5} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Ersetze zurück

= - \frac{1}{5} \frac{ln \frac{25 - t ^{2}}{5} + 1}{2} - \frac{ln \frac{25 - t ^{2}}{5} - 1}{2}

Vereinfache

- \frac{1}{5} \frac{ln \frac{25 - t ^{2}}{5} + 1}{2} - \frac{ln \frac{25 - t ^{2}}{5} - 1}{2} : - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1)

= - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{10} (ln \frac{25 - t ^{2}}{5} + 1 - ln \frac{25 - t ^{2}}{5} - 1) + C

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