integral of (sqrt(1-x))/(sqrt(2-x))

Lösung

\int \frac{1 - x}{2 - x} d x

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

Schritte zur Lösung

\int \frac{1 - x}{2 - x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= - \int \frac{v - 1}{v} d v

Wende die partielle Integration an

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

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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