integral of (8x)/(x^3+x^2-x-1)

Lösung

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

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

Schritte zur Lösung

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

Entferne die Konstante:

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

= 8 \cdot \int \frac{x}{x ^{3} + x ^{2} - x - 1} d x

Ermittle den Partialbruch von

\frac{x}{x ^{3} + x ^{2} - x - 1} : - \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )}

= 8 \cdot \int - \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )} 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

= 8 (- \int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{2 ( x + 1 ) ^{2}} d x + \int \frac{1}{4 ( x - 1 )} d x)

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

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

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

= 8 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 8 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + C

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