integral of (x^3)/(x+4)

Lösung

\int \frac{x ^{3}}{x + 4} d x

\frac{x ^{3}}{3} - 2 x^{2} + 16 x - 64 ln ∣ x + 4 ∣ + C

Schritte zur Lösung

\int \frac{x ^{3}}{x + 4} d x

Lange Division

\frac{x ^{3}}{x + 4} : x^{2} - 4 x + 16 - \frac{64}{x + 4}

= \int x^{2} - 4 x + 16 - \frac{64}{x + 4} 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 4 x d x + \int 16 d x - \int \frac{64}{x + 4} d x

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

\int 4 x d x = 2 x^{2}

\int 16 d x = 16 x

\int \frac{64}{x + 4} d x = 64 ln ∣ x + 4 ∣

= \frac{x ^{3}}{3} - 2 x^{2} + 16 x - 64 ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - 2 x^{2} + 16 x - 64 ln ∣ x + 4 ∣ + C

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

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