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Beliebt Rechnen >

integral of 5/((x^2-1)^2)

Lösung

\int \frac{5}{( x ^{2} - 1 ) ^{2}} d x

Lösung

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

Schritte zur Lösung

\int \frac{5}{( x ^{2} - 1 ) ^{2}} d x

Entferne die Konstante:

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

= 5 \cdot \int \frac{1}{( x ^{2} - 1 ) ^{2}} d x

(x^{2} - 1)^{2} = x^{4} - 2 x^{2} + 1

= 5 \cdot \int \frac{1}{x ^{4} - 2 x ^{2} + 1} d x

Ermittle den Partialbruch von

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

= 5 \cdot \int \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{2}} - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 1 ) ^{2}} d x

Wende die Summenregel an:

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

= 5 (\int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{4 ( x + 1 ) ^{2}} d x - \int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{4 ( x - 1 ) ^{2}} d x)

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

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

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

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

= 5 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )})

Füge eine Konstante zur Lösung hinzu

= 5 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) + C

Graph

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