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integral of 1/((x^2-2x+5)^2)

Lösung

\int \frac{1}{( x ^{2} - 2 x + 5 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{1}{( x ^{2} - 2 x + 5 ) ^{2}} d x

Vervollständige das Quadrat

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

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

Wende U-Substitution an

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

Wende integrale Substitution an

= \int \frac{1}{8 ( v ^{2} + 1 ) ^{2}} d v

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Trigonometrische Substitution anwenden

= \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{8} \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

= \frac{1}{8} \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

= \frac{1}{8} \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)

= \frac{1}{8} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Ersetze zurück

= \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x - 1}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 1}{2})))

Vereinfache

\frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x - 1}{2}) + \frac{1}{2} sin (2 arctan (\frac{x - 1}{2}))) : \frac{1}{16} (arctan (\frac{1}{2} (x - 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 1))))

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

Füge eine Konstante zur Lösung hinzu

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

f (x) = - x^{4} + 18 x^{2} - 11

\frac{d}{d x} ((x^{2} + 7) 3 x^{2} + 2)

\int - 4 sin (4 t) + 1 d t

x \to + 4 lim (9 x + ∣ x - 4 ∣)

d er i v a t i v e - sin^{3} (x)