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Beliebt Rechnen >

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

Lösung

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

Lösung

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

Schritte zur Lösung

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

Wende U-Substitution an

= \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 = x + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

Beliebte Beispiele

integral of (x^2)/((x-3)^5)

\int \frac{x ^{2}}{( x - 3 ) ^{5}} d x

integral of 3x(x^2+8)^2

\int 3 x (x^{2} + 8)^{2} d x

integral of x^2(x^3+8)^{10}

\int x^{2} (x^{3} + 8)^{10} d x

integral of 1/((x+3)sqrt(x+1))

\int \frac{1}{( x + 3 ) x + 1} d x

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

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