integral of (x^6-x^3+16)/(x^4+4x^2)

Lösung

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

\frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C

Schritte zur Lösung

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

Ermittle den Partialbruch von

\frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} : x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{x ^{2} + 4}

= \int x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{x ^{2} + 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 d x + \int \frac{4}{x ^{2}} d x + \int \frac{- x + 12}{x ^{2} + 4} d x

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

\int 4 d x = 4 x

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

\int \frac{- x + 12}{x ^{2} + 4} d x = - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

- 6 y^{^{'}} + 18 x^{2} = 0

(t^{3}) d x + (x t^{2} + x^{3}) d t = 0

\int cos (x) - cos^{3} (x) d x

\frac{d}{d x} (- 4 x^{- 3})