integral of 1/((x-1)sqrt(15+2x-x^2))

Lösung

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

- \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

= - \frac{1}{4} \cdot \int \frac{1}{- w ^{2} + 1} d w

Nutze das gemeinsame Integral :

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

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

Ersetze zurück

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

Vereinfache

- \frac{1}{4} \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- ( x - 1 ) ^{2} + 16}{4} - 1}{2} : - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1)

= - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{8} (ln \frac{- x ^{2} + 2 x + 15}{4} + 1 - ln \frac{- x ^{2} + 2 x + 15}{4} - 1) + C

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