integral of (x^6-x^3+1)/(x^4+9x^2)

Lösung

\int \frac{x ^{6} - x ^{3} + 1}{x ^{4} + 9 x ^{2}} d x

\frac{x ^{3}}{3} - 9 x - \frac{1}{2} ln x^{2} + 9 + \frac{728}{27} arctan (\frac{x}{3}) - \frac{1}{9 x} + C

Schritte zur Lösung

\int \frac{x ^{6} - x ^{3} + 1}{x ^{4} + 9 x ^{2}} d x

Faktorisiere

\frac{x ^{6} - x ^{3} + 1}{x ^{4} + 9 x ^{2}}

= \int \frac{x ^{6} - x ^{3} + 1}{x ^{2} ( x ^{2} + 9 )} d x

Multipliziere aus

\frac{x ^{6} - x ^{3} + 1}{x ^{2} ( x ^{2} + 9 )} : \frac{x ^{6}}{x ^{2} ( x ^{2} + 9 )} - \frac{x ^{3}}{x ^{2} ( x ^{2} + 9 )} + \frac{1}{x ^{2} ( x ^{2} + 9 )}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{6}}{x ^{2} ( x ^{2} + 9 )} d x - \int \frac{x ^{3}}{x ^{2} ( x ^{2} + 9 )} d x + \int \frac{1}{x ^{2} ( x ^{2} + 9 )} d x

\int \frac{x ^{6}}{x ^{2} ( x ^{2} + 9 )} d x = 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3}))

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

\int \frac{1}{x ^{2} ( x ^{2} + 9 )} d x = - \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})

= 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) - \frac{1}{2} ln x^{2} + 9 - \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})

Vereinfache

27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3})) - \frac{1}{2} ln x^{2} + 9 - \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}) : \frac{x ^{3}}{3} - 9 x - \frac{1}{2} ln x^{2} + 9 + \frac{728}{27} arctan (\frac{x}{3}) - \frac{1}{9 x}

= \frac{x ^{3}}{3} - 9 x - \frac{1}{2} ln x^{2} + 9 + \frac{728}{27} arctan (\frac{x}{3}) - \frac{1}{9 x}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} - 9 x - \frac{1}{2} ln x^{2} + 9 + \frac{728}{27} arctan (\frac{x}{3}) - \frac{1}{9 x} + C

x \to \infty lim (\frac{lo g _{4} ( x )}{lo g _{9} ( x )})

\int_{1}^{5} x - 1 d x

\int \frac{3}{7} x^{- \frac{1}{7}} d x

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

k = 0 \sum \infty k! 2^{k} x^{k}