integral of (x^3)/(x^4sqrt(16-x^2))

Lösung

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

- \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

Schritte zur Lösung

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

Vereinfache

\frac{x ^{3}}{x ^{4} 16 - x ^{2}} : \frac{1}{x - x ^{2} + 16}

= \int \frac{1}{x - x ^{2} + 16} d x

Wende U-Substitution an

= \int - \frac{1}{- u ^{2} + 16} d u

Entferne die Konstante:

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

= - \int \frac{1}{- u ^{2} + 16} d u

Wende integrale Substitution an

= - \int \frac{1}{4 ( - v ^{2} + 1 )} d v

Entferne die Konstante:

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

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

Ersetze zurück

= - \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2}

Vereinfache

- \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2} : - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

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