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Beliebt Rechnen >

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

Lösung

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

Lösung

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

\frac{1}{( - u ^{2} + 1 ) ^{2}} : \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}}

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

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

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

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

Setze in

u = 1 - x

ein

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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