integral of (2x+1)/(x^2-2x+5)

Lösung

\int \frac{2 x + 1}{x ^{2} - 2 x + 5} d x

ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) + \frac{1}{2} arctan (\frac{x - 1}{2}) + C

Schritte zur Lösung

\int \frac{2 x + 1}{x ^{2} - 2 x + 5} d x

Multipliziere aus

\frac{2 x + 1}{x ^{2} - 2 x + 5} : \frac{2 x}{x ^{2} - 2 x + 5} + \frac{1}{x ^{2} - 2 x + 5}

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{2 x}{x ^{2} - 2 x + 5} d x + \int \frac{1}{x ^{2} - 2 x + 5} d x

\int \frac{2 x}{x ^{2} - 2 x + 5} d x = ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1))

\int \frac{1}{x ^{2} - 2 x + 5} d x = \frac{1}{2} arctan (\frac{x - 1}{2})

= ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) + \frac{1}{2} arctan (\frac{x - 1}{2})

Füge eine Konstante zur Lösung hinzu

= ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) + \frac{1}{2} arctan (\frac{x - 1}{2}) + C

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