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

Lösung

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

\frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{x ^{3}}{x ^{2} - 1} d x = \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1)

\int \frac{3}{x ^{2} - 1} d x = - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) - 3 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

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