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integral of (x^2-4x+4)/(sqrt(x^2-4x+29))

Lösung

\int \frac{x ^{2} - 4 x + 4}{x ^{2} - 4 x + 29} d x

25 (- ln (\frac{1}{5} x - 2 + x^{2} - 4 x + 29) + \frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5})) + C

Schritte zur Lösung

\int \frac{x ^{2} - 4 x + 4}{x ^{2} - 4 x + 29} d x

Vervollständige das Quadrat

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

= \int \frac{( x - 2 ) ^{2}}{x ^{2} - 4 x + 29} d x

Wende U-Substitution an

= \int \frac{u ^{2}}{u ^{2} + 25} d u

Trigonometrische Substitution anwenden

= \int 25 tan^{2} (v) sec (v) d v

Entferne die Konstante:

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

= 25 \cdot \int tan^{2} (v) sec (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 25 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Multipliziere aus

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 25 \cdot \int - sec (v) + sec^{3} (v) d v

Wende die Summenregel an:

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

= 25 (- \int sec (v) d v + \int sec^{3} (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 25 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= 25 (- ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2))) + \frac{1}{2} sec (arctan (\frac{1}{5} (x - 2))) tan (arctan (\frac{1}{5} (x - 2))) + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2))))

Vereinfache

25 (- ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2))) + \frac{1}{2} sec (arctan (\frac{1}{5} (x - 2))) tan (arctan (\frac{1}{5} (x - 2))) + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2)))) : 25 (- ln (\frac{1}{5} x - 2 + x^{2} - 4 x + 29) + \frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5}))

= 25 (- ln (\frac{1}{5} x - 2 + x^{2} - 4 x + 29) + \frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5}))

Füge eine Konstante zur Lösung hinzu

= 25 (- ln (\frac{1}{5} x - 2 + x^{2} - 4 x + 29) + \frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5})) + C

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