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

Lösung

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

\frac{x ^{2}}{2} + x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{8}{3} ln ∣ x - 2 ∣ + C

Schritte zur Lösung

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

Lange Division

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

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

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

\int 1 d x = x

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

= \frac{x ^{2}}{2} + x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{8}{3} ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} + x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{8}{3} ln ∣ x - 2 ∣ + C

d er i v a t i v e y = x + x^{π}

t an g e n t f (x) = \frac{1}{4 + 3 x}, at x = 2

y = - x^{- 1} + \frac{3}{5} x + \frac{1}{2}

\int (t^{2} + 2 t + 1)^{4} (t + 1) d t

\int_{0}^{5} ∣ 9 - 3 t ∣ d t