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integral of 1/((x^2-6x+11)^2)

Lösung

\int \frac{1}{( x ^{2} - 6 x + 11 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{1}{( x ^{2} - 6 x + 11 ) ^{2}} d x

Vervollständige das Quadrat

x^{2} - 6 x + 11 : (x - 3)^{2} + 2

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Faktorisiere

\frac{1}{( 2 v ^{2} + 2 ) ^{2}}

= 2 \cdot \int \frac{1}{4 ( v ^{2} + 1 ) ^{2}} d v

Entferne die Konstante:

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

= 2 \frac{1}{4} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Trigonometrische Substitution anwenden

= 2 \frac{1}{4} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 2 \frac{1}{4} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Entferne die Konstante:

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

= 2 \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \frac{1}{4} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= 2 \frac{1}{4} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e y = - 8^{5 x^{2} - 4}

x \to - 1 lim (3 x^{4} - 2 x^{3} + 4 x)

\frac{\partial}{\partial x} (ln (1 + z) + x y^{2} + z)

in v erse l a pl a ce \frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}

d er i v a t i v e y = x^{4} cos (4 x^{2})