integral of (sqrt(225-x^2))/x

Lösung

\int \frac{225 - x ^{2}}{x} d x

- \frac{15}{2} ln \frac{225 - x ^{2}}{15} + 1 + \frac{15}{2} ln \frac{225 - x ^{2}}{15} - 1 + 225 - x^{2} + C

Schritte zur Lösung

\int \frac{225 - x ^{2}}{x} d x

Wende U-Substitution an

= \int - \frac{u ^{2}}{- u ^{2} + 225} d u

Entferne die Konstante:

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

= - \int \frac{u ^{2}}{- u ^{2} + 225} d u

Wende integrale Substitution an

= - \int \frac{15 v ^{2}}{- v ^{2} + 1} d v

Entferne die Konstante:

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

= - 15 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - 15 \cdot \int \frac{1}{- v ^{2} + 1} - 1 d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 15 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v)

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

\int 1 d v = v

= - 15 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

Ersetze zurück

= - 15 \frac{ln \frac{225 - x ^{2}}{15} + 1}{2} - \frac{ln \frac{225 - x ^{2}}{15} - 1}{2} - \frac{225 - x ^{2}}{15}

Vereinfache

- 15 \frac{ln \frac{225 - x ^{2}}{15} + 1}{2} - \frac{ln \frac{225 - x ^{2}}{15} - 1}{2} - \frac{225 - x ^{2}}{15} : - \frac{15}{2} ln \frac{225 - x ^{2}}{15} + 1 + \frac{15}{2} ln \frac{225 - x ^{2}}{15} - 1 + 225 - x^{2}

= - \frac{15}{2} ln \frac{225 - x ^{2}}{15} + 1 + \frac{15}{2} ln \frac{225 - x ^{2}}{15} - 1 + 225 - x^{2}

Füge eine Konstante zur Lösung hinzu

= - \frac{15}{2} ln \frac{225 - x ^{2}}{15} + 1 + \frac{15}{2} ln \frac{225 - x ^{2}}{15} - 1 + 225 - x^{2} + C

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