integral of x/(x^2+6x+18)

Lösung

\int \frac{x}{x ^{2} + 6 x + 18} d x

\frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3)) + C

Schritte zur Lösung

\int \frac{x}{x ^{2} + 6 x + 18} d x

Vervollständige das Quadrat

x^{2} + 6 x + 18 : (x + 3)^{2} + 9

= \int \frac{x}{( x + 3 ) ^{2} + 9} d x

Wende U-Substitution an

= \int \frac{u - 3}{u ^{2} + 9} d u

Multipliziere aus

\frac{u - 3}{u ^{2} + 9} : \frac{u}{u ^{2} + 9} - \frac{3}{u ^{2} + 9}

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} + 9} d u - \int \frac{3}{u ^{2} + 9} d u

\int \frac{u}{u ^{2} + 9} d u = \frac{1}{2} ln u^{2} + 9

\int \frac{3}{u ^{2} + 9} d u = arctan (\frac{u}{3})

= \frac{1}{2} ln u^{2} + 9 - arctan (\frac{u}{3})

Setze in

u = x + 3

ein

= \frac{1}{2} ln (x + 3)^{2} + 9 - arctan (\frac{x + 3}{3})

Vereinfache

\frac{1}{2} ln (x + 3)^{2} + 9 - arctan (\frac{x + 3}{3}) : \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3))

= \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3)) + C

in v erse l a pl a ce \frac{s + 1}{s ( s ^{2} + 4 s + 4 )}

\frac{d}{d x} (x csc (x))

d er i v a t i v e x^{3} + 2 x

\int 9 sin (x) cos (x) d x

(2 e^{y} - x) y^{^{'}} = 1