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Beliebt Rechnen >

integral 1/([(1-x^2)(1+x)])

Lösung

integral

\frac{1}{[ ( 1 - x ^{2} ) ( 1 + x ) ]}

Lösung

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

Schritte zur Lösung

\int \frac{1}{[ ( 1 - x ^{2} ) ( 1 + x ) ]} d x

Vereinfache

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

Wende U-Substitution an

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

Faktorisiere

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

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

Setze in

u = 1 + x

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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