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

Lösung

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

\frac{x ^{3}}{3} - x^{2} + 4 x - 7 ln ∣ x + 2 ∣ + C

Schritte zur Lösung

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

Lange Division

\frac{x ^{3} + 1}{x + 2} : x^{2} - 2 x + 4 - \frac{7}{x + 2}

= \int x^{2} - 2 x + 4 - \frac{7}{x + 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

= \int x^{2} d x - \int 2 x d x + \int 4 d x - \int \frac{7}{x + 2} d x

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

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

\int 4 d x = 4 x

\int \frac{7}{x + 2} d x = 7 ln ∣ x + 2 ∣

= \frac{x ^{3}}{3} - x^{2} + 4 x - 7 ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - x^{2} + 4 x - 7 ln ∣ x + 2 ∣ + C

(e^{- x})^{^{'}}

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

n = 0 \sum \infty (\frac{2 π}{3})^{n}

x \to \frac{π}{2} lim (sin (x - a))

\int - x + 3 d x