integral of (-9x+27)/((x^2+2x+4)^2)

Lösung

\int \frac{- 9 x + 27}{( x ^{2} + 2 x + 4 ) ^{2}} d x

\frac{9}{2 ( x ^{2} + 2 x + 4 )} + 23 arctan (\frac{1}{3} (x + 1)) + 3 sin (2 arctan (\frac{1}{3} (x + 1))) + C

Schritte zur Lösung

\int \frac{- 9 x + 27}{( x ^{2} + 2 x + 4 ) ^{2}} d x

Vervollständige das Quadrat

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

= \int \frac{- 9 x + 27}{( ( x + 1 ) ^{2} + 3 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{- 9 u + 36}{( u ^{2} + 3 ) ^{2}} d u

Multipliziere aus

\frac{- 9 u + 36}{( u ^{2} + 3 ) ^{2}} : - \frac{9 u}{( u ^{2} + 3 ) ^{2}} + \frac{36}{( 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{9 u}{( u ^{2} + 3 ) ^{2}} d u + \int \frac{36}{( u ^{2} + 3 ) ^{2}} d u

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

\int \frac{36}{( u ^{2} + 3 ) ^{2}} d u = 23 arctan (\frac{u}{3}) + 3 sin (2 arctan (\frac{u}{3}))

= - (- \frac{9}{2 ( u ^{2} + 3 )}) + 23 arctan (\frac{u}{3}) + 3 sin (2 arctan (\frac{u}{3}))

Setze in

u = x + 1

ein

= - - \frac{9}{2 ( ( x + 1 ) ^{2} + 3 )} + 23 arctan (\frac{x + 1}{3}) + 3 sin (2 arctan (\frac{x + 1}{3}))

Vereinfache

- - \frac{9}{2 ( ( x + 1 ) ^{2} + 3 )} + 23 arctan (\frac{x + 1}{3}) + 3 sin (2 arctan (\frac{x + 1}{3})) : \frac{9}{2 ( x ^{2} + 2 x + 4 )} + 23 arctan (\frac{1}{3} (x + 1)) + 3 sin (2 arctan (\frac{1}{3} (x + 1)))

= \frac{9}{2 ( x ^{2} + 2 x + 4 )} + 23 arctan (\frac{1}{3} (x + 1)) + 3 sin (2 arctan (\frac{1}{3} (x + 1)))

Füge eine Konstante zur Lösung hinzu

= \frac{9}{2 ( x ^{2} + 2 x + 4 )} + 23 arctan (\frac{1}{3} (x + 1)) + 3 sin (2 arctan (\frac{1}{3} (x + 1))) + C

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