integral of 1/((x+1)sqrt(1-x))

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

= - 2 \cdot \frac{1}{2} \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}

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

Ersetze zurück

= - 2 \cdot \frac{1}{2} \frac{ln \frac{- ( x + 1 ) + 2}{2} + 1}{2} - \frac{ln \frac{- ( x + 1 ) + 2}{2} - 1}{2}

Vereinfache

- 2 \cdot \frac{1}{2} \frac{ln \frac{- ( x + 1 ) + 2}{2} + 1}{2} - \frac{ln \frac{- ( x + 1 ) + 2}{2} - 1}{2} : - 2 (\frac{1}{2} ln \frac{- x + 1}{2} + 1 - \frac{1}{2} ln \frac{- x + 1}{2} - 1)

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

Füge eine Konstante zur Lösung hinzu

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

\int x^{5} + 5 x^{7} d x

\int 2 arctan (4 y) d y

\int x^{5} \cdot e^{- x^{3}} d x

\int e^{- x^{9}} d x

\int \frac{1}{( x - 3 ) ( x + 1 )} d x