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

Lösung

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

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

Schritte zur Lösung

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

Lange Division

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

= \int x^{2} - x + 2 - \frac{2}{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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int x d x = \frac{x ^{2}}{2}

\int 2 d x = 2 x

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

Füge eine Konstante zur Lösung hinzu

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

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