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Beliebt Rechnen >

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

Lösung

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

Lösung

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

Schritte zur Lösung

\int \frac{1 - x}{1 + x} \cdot \frac{1}{x} d x

Vereinfache

\frac{1 - x}{1 + x} : \frac{1 - x}{1 + x}

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

Setze in

u = \frac{1 - x}{1 + x}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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