integral of (x^3-x^2+x-1)/(x^2+9)

Lösung

\int \frac{x ^{3} - x ^{2} + x - 1}{x ^{2} + 9} d x

\frac{x ^{2}}{2} - x - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3}) + C

Schritte zur Lösung

\int \frac{x ^{3} - x ^{2} + x - 1}{x ^{2} + 9} d x

Lange Division

\frac{x ^{3} - x ^{2} + x - 1}{x ^{2} + 9} : x - 1 + \frac{- 8 x + 8}{x ^{2} + 9}

= \int x - 1 + \frac{- 8 x + 8}{x ^{2} + 9} 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{- 8 x + 8}{x ^{2} + 9} d x

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

\int 1 d x = x

\int \frac{- 8 x + 8}{x ^{2} + 9} d x = - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3})

= \frac{x ^{2}}{2} - x - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} - x - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3}) + C

d er i v a t i v e \frac{x + 1}{x ^{2}}

x \to 2 lim (lo g_{4} (x - 2))

\frac{d y}{d x} + 3 x^{2} y = x^{2}

\frac{d y}{d x} = 10 x e^{5 y}

d er i v a t i v e f (x) = \frac{2}{x}