integral of 1/(sqrt(x)(1-\sqrt{1-x))}

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{u - 1}{u - u ^{2} + 2 u} : \frac{1}{- u ^{2} + 2 u} - \frac{1}{u - u ^{2} + 2 u}

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

Wende die Summenregel an:

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

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

\int \frac{1}{- u ^{2} + 2 u} d u = arcsin (u - 1)

\int \frac{1}{u - u ^{2} + 2 u} d u = - \frac{2}{tan ( \frac{1}{2} arcsin ( u - 1 ) ) + 1}

= 2 (arcsin (u - 1) - (- \frac{2}{tan ( \frac{1}{2} arcsin ( u - 1 ) ) + 1}))

Setze in

u = 1 - 1 - x

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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