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

Lösung

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

- \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2}) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

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

= \int \frac{x}{( ( x + 1 ) ^{2} + 1 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u - 1}{( u ^{2} + 1 ) ^{2}} d u

Multipliziere aus

\frac{u - 1}{( u ^{2} + 1 ) ^{2}} : \frac{u}{( u ^{2} + 1 ) ^{2}} - \frac{1}{( u ^{2} + 1 ) ^{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{u}{( u ^{2} + 1 ) ^{2}} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

\int \frac{u}{( u ^{2} + 1 ) ^{2}} d u = - \frac{1}{2 ( u ^{2} + 1 )}

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

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

Setze in

u = x + 1

ein

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

Vereinfache

- \frac{1}{2 ( ( x + 1 ) ^{2} + 1 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{1 + ( x + 1 ) ^{2}}) : - \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2})

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

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2}) + C

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