integral of 1/((x^2+1)(x^2+x+1))

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

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