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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

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

Wende U-Substitution an

= \int \frac{2 u + 1}{u ^{2} + 1} d u

Multipliziere aus

\frac{2 u + 1}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} + \frac{1}{u ^{2} + 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{2 u}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u

\int \frac{2 u}{u ^{2} + 1} d u = ln u^{2} + 1

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

= ln u^{2} + 1 + arctan (u)

Setze in

u = x - 1

ein

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

Multipliziere aus

(x - 1)^{2} + 1 : x^{2} - 2 x + 2

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

Füge eine Konstante zur Lösung hinzu

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

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