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

Lösung

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

- \frac{1}{2} ln 16 x^{2} + 8 x + 8 + \frac{3}{7} arctan (\frac{1}{7} (4 x + 1)) + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{1}{2 x ^{2} + x + 1} d x = \frac{2}{7} arctan (\frac{4 x + 1}{7})

\int \frac{2 x}{2 x ^{2} + x + 1} d x = \frac{1}{2} ln 16 x^{2} + 8 x + 8 - \frac{1}{7} arctan (\frac{4 x + 1}{7})

= \frac{2}{7} arctan (\frac{4 x + 1}{7}) - (\frac{1}{2} ln 16 x^{2} + 8 x + 8 - \frac{1}{7} arctan (\frac{4 x + 1}{7}))

Vereinfache

\frac{2}{7} arctan (\frac{4 x + 1}{7}) - (\frac{1}{2} ln 16 x^{2} + 8 x + 8 - \frac{1}{7} arctan (\frac{4 x + 1}{7})) : - \frac{1}{2} ln 16 x^{2} + 8 x + 8 + \frac{3}{7} arctan (\frac{1}{7} (4 x + 1))

= - \frac{1}{2} ln 16 x^{2} + 8 x + 8 + \frac{3}{7} arctan (\frac{1}{7} (4 x + 1))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln 16 x^{2} + 8 x + 8 + \frac{3}{7} arctan (\frac{1}{7} (4 x + 1)) + C

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