integral of (8x-24)/((x^2+2x+4)^2)

Lösung

\int \frac{8 x - 24}{( x ^{2} + 2 x + 4 ) ^{2}} d x

- \frac{4}{x ^{2} + 2 x + 4} - \frac{16}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

Schritte zur Lösung

\int \frac{8 x - 24}{( x ^{2} + 2 x + 4 ) ^{2}} d x

Vervollständige das Quadrat

x^{2} + 2 x + 4 : (x + 1)^{2} + 3

= \int \frac{8 x - 24}{( ( x + 1 ) ^{2} + 3 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{8 u - 32}{( u ^{2} + 3 ) ^{2}} d u

Multipliziere aus

\frac{8 u - 32}{( u ^{2} + 3 ) ^{2}} : \frac{8 u}{( u ^{2} + 3 ) ^{2}} - \frac{32}{( u ^{2} + 3 ) ^{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{8 u}{( u ^{2} + 3 ) ^{2}} d u - \int \frac{32}{( u ^{2} + 3 ) ^{2}} d u

\int \frac{8 u}{( u ^{2} + 3 ) ^{2}} d u = - \frac{4}{u ^{2} + 3}

\int \frac{32}{( u ^{2} + 3 ) ^{2}} d u = \frac{16}{3 3} (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3})))

= - \frac{4}{u ^{2} + 3} - \frac{16}{3 3} (arctan (\frac{u}{3}) + \frac{1}{2} sin (2 arctan (\frac{u}{3})))

Setze in

u = x + 1

ein

= - \frac{4}{( x + 1 ) ^{2} + 3} - \frac{16}{3 3} (arctan (\frac{x + 1}{3}) + \frac{1}{2} sin (2 arctan (\frac{x + 1}{3})))

Vereinfache

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

= - \frac{4}{x ^{2} + 2 x + 4} - \frac{16}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1))))

Füge eine Konstante zur Lösung hinzu

= - \frac{4}{x ^{2} + 2 x + 4} - \frac{16}{3 3} (arctan (\frac{1}{3} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{3} (x + 1)))) + C

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