integral of (2x+3)/(x^2+2x+3)

Lösung

\int \frac{2 x + 3}{x ^{2} + 2 x + 3} d x

ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

Schritte zur Lösung

\int \frac{2 x + 3}{x ^{2} + 2 x + 3} d x

Vervollständige das Quadrat

x^{2} + 2 x + 3 : (x + 1)^{2} + 2

= \int \frac{2 x + 3}{( x + 1 ) ^{2} + 2} d x

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

Setze in

u = x + 1

ein

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

Vereinfache

ln (x + 1)^{2} + 2 + \frac{1}{2} arctan (\frac{x + 1}{2}) : ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

= ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C