integral of (6x+15)/(x^2+4x+20)

Lösung

\int \frac{6 x + 15}{x ^{2} + 4 x + 20} d x

3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2)) + C

Schritte zur Lösung

\int \frac{6 x + 15}{x ^{2} + 4 x + 20} d x

Vervollständige das Quadrat

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int \frac{6 x + 15}{( x + 2 ) ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{6 u + 3}{u ^{2} + 16} d u

Multipliziere aus

\frac{6 u + 3}{u ^{2} + 16} : \frac{6 u}{u ^{2} + 16} + \frac{3}{u ^{2} + 16}

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{6 u}{u ^{2} + 16} d u + \int \frac{3}{u ^{2} + 16} d u

\int \frac{6 u}{u ^{2} + 16} d u = 3 ln u^{2} + 16

\int \frac{3}{u ^{2} + 16} d u = \frac{3}{4} arctan (\frac{u}{4})

= 3 ln u^{2} + 16 + \frac{3}{4} arctan (\frac{u}{4})

Setze in

u = x + 2

ein

= 3 ln (x + 2)^{2} + 16 + \frac{3}{4} arctan (\frac{x + 2}{4})

Vereinfache

3 ln (x + 2)^{2} + 16 + \frac{3}{4} arctan (\frac{x + 2}{4}) : 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2))

= 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2))

Füge eine Konstante zur Lösung hinzu

= 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2)) + C

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