integral of ((2x+6))/(x^2+4x+8)

Lösung

\int \frac{( 2 x + 6 )}{x ^{2} + 4 x + 8} d x

ln x^{2} + 4 x + 8 + arctan (\frac{1}{2} (x + 2)) + C

Schritte zur Lösung

\int \frac{2 x + 6}{x ^{2} + 4 x + 8} d x

Vervollständige das Quadrat

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

= \int \frac{2 x + 6}{( x + 2 ) ^{2} + 4} d x

Wende U-Substitution an

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

Multipliziere aus

\frac{2 u + 2}{u ^{2} + 4} : \frac{2 u}{u ^{2} + 4} + \frac{2}{u ^{2} + 4}

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

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

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

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

Setze in

u = x + 2

ein

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

Vereinfache

ln (x + 2)^{2} + 4 + arctan (\frac{x + 2}{2}) : ln x^{2} + 4 x + 8 + arctan (\frac{1}{2} (x + 2))

= ln x^{2} + 4 x + 8 + arctan (\frac{1}{2} (x + 2))

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 4 x + 8 + arctan (\frac{1}{2} (x + 2)) + C

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