integral of sqrt(4x^2-4x+5)

Lösung

\int 4 x^{2} - 4 x + 5 d x

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

Schritte zur Lösung

\int 4 x^{2} - 4 x + 5 d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int 2 u^{2} + 1 d u

Entferne die Konstante:

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

= 2 \cdot \int u^{2} + 1 d u

Trigonometrische Substitution anwenden

= 2 \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

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

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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