integral of (x+2)/(x^2-4x+5)

Lösung

\int \frac{x + 2}{x ^{2} - 4 x + 5} d x

\frac{1}{2} ln x^{2} - 4 x + 5 + 4 arctan (x - 2) + C

Schritte zur Lösung

\int \frac{x + 2}{x ^{2} - 4 x + 5} d x

Vervollständige das Quadrat

x^{2} - 4 x + 5 : (x - 2)^{2} + 1

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

Wende U-Substitution an

= \int \frac{u + 4}{u ^{2} + 1} d u

Multipliziere aus

\frac{u + 4}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{4}{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{u}{u ^{2} + 1} d u + \int \frac{4}{u ^{2} + 1} d u

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

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

= \frac{1}{2} ln u^{2} + 1 + 4 arctan (u)

Setze in

u = x - 2

ein

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

Vereinfache

\frac{1}{2} ln (x - 2)^{2} + 1 + 4 arctan (x - 2) : \frac{1}{2} ln x^{2} - 4 x + 5 + 4 arctan (x - 2)

= \frac{1}{2} ln x^{2} - 4 x + 5 + 4 arctan (x - 2)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 4 x + 5 + 4 arctan (x - 2) + C

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