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

Lösung

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

23 arctan (\frac{1}{3} (2 x + 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{3 + x}{x ^{2} + x + 1} d x

Multipliziere aus

\frac{3 + x}{x ^{2} + x + 1} : \frac{3}{x ^{2} + x + 1} + \frac{x}{x ^{2} + x + 1}

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

\int \frac{3}{x ^{2} + x + 1} d x = 23 arctan (\frac{1}{3} (2 x + 1))

\int \frac{x}{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})

= 23 arctan (\frac{1}{3} (2 x + 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

= 23 arctan (\frac{1}{3} (2 x + 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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