integral ((1-x))/((x^2-x+4))

Lösung

integral

\frac{( 1 - x )}{( x ^{2} - x + 4 )}

- \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1)) + C

Schritte zur Lösung

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

Multipliziere aus

\frac{1 - x}{x ^{2} - x + 4} : \frac{1}{x ^{2} - x + 4} - \frac{x}{x ^{2} - x + 4}

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

\int \frac{1}{x ^{2} - x + 4} d x = \frac{2}{15} arctan (\frac{2 x - 1}{15})

\int \frac{x}{x ^{2} - x + 4} d x = \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15})

= \frac{2}{15} arctan (\frac{2 x - 1}{15}) - (\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15}))

Vereinfache

\frac{2}{15} arctan (\frac{2 x - 1}{15}) - (\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15})) : - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1))

= - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1)) + C

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