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Beliebt Rechnen >

integral of (x-2)/((2x^2+2x+1)^2)

Lösung

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

Lösung

2 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 2 )} - \frac{5}{8} (2 arctan (2 x + 1) + sin (2 arctan (2 x + 1)))) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int \frac{2 ( 2 u - 5 )}{( 4 u ^{2} + 1 ) ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u - 5}{( 4 u ^{2} + 1 ) ^{2}} d u

Multipliziere aus

\frac{2 u - 5}{( 4 u ^{2} + 1 ) ^{2}} : \frac{2 u}{( 4 u ^{2} + 1 ) ^{2}} - \frac{5}{( 4 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

= 2 (\int \frac{2 u}{( 4 u ^{2} + 1 ) ^{2}} d u - \int \frac{5}{( 4 u ^{2} + 1 ) ^{2}} d u)

\int \frac{2 u}{( 4 u ^{2} + 1 ) ^{2}} d u = - \frac{1}{4 ( 4 u ^{2} + 1 )}

\int \frac{5}{( 4 u ^{2} + 1 ) ^{2}} d u = \frac{5}{4} (arctan (2 u) + \frac{1}{2} sin (2 arctan (2 u)))

= 2 (- \frac{1}{4 ( 4 u ^{2} + 1 )} - \frac{5}{4} (arctan (2 u) + \frac{1}{2} sin (2 arctan (2 u))))

Setze in

u = x + \frac{1}{2}

ein

= 2 - \frac{1}{4 ( 4 ( x + \frac{1}{2} ) ^{2} + 1 )} - \frac{5}{4} (arctan (2 (x + \frac{1}{2})) + \frac{1}{2} sin (2 arctan (2 (x + \frac{1}{2}))))

Vereinfache

2 - \frac{1}{4 ( 4 ( x + \frac{1}{2} ) ^{2} + 1 )} - \frac{5}{4} (arctan (2 (x + \frac{1}{2})) + \frac{1}{2} sin (2 arctan (2 (x + \frac{1}{2})))) : 2 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 2 )} - \frac{5}{8} (2 arctan (2 x + 1) + sin (2 arctan (2 x + 1))))

= 2 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 2 )} - \frac{5}{8} (2 arctan (2 x + 1) + sin (2 arctan (2 x + 1))))

Füge eine Konstante zur Lösung hinzu

= 2 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 2 )} - \frac{5}{8} (2 arctan (2 x + 1) + sin (2 arctan (2 x + 1)))) + C

Graph

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