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

Lösung

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

\frac{1}{2} ln x^{2} - 2 x + 2 + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Vereinfache

= \frac{1}{2} ln x^{2} - 2 x + 2

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 2 x + 2 + C