integral of ((x^4+9x^2+x+2))/((x^2+9))

Lösung

\int \frac{( x ^{4} + 9 x ^{2} + x + 2 )}{( x ^{2} + 9 )} d x

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

Schritte zur Lösung

\int \frac{x ^{4} + 9 x ^{2} + x + 2}{x ^{2} + 9} d x

Multipliziere aus

\frac{x ^{4} + 9 x ^{2} + x + 2}{x ^{2} + 9} : \frac{x ^{4}}{x ^{2} + 9} + \frac{9 x ^{2}}{x ^{2} + 9} + \frac{x}{x ^{2} + 9} + \frac{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 ^{4}}{x ^{2} + 9} d x + \int \frac{9 x ^{2}}{x ^{2} + 9} d x + \int \frac{x}{x ^{2} + 9} d x + \int \frac{2}{x ^{2} + 9} d x

\int \frac{x ^{4}}{x ^{2} + 9} d x = 27 (\frac{x ^{3}}{81} - \frac{x}{3} + arctan (\frac{x}{3}))

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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