integral of sqrt(4x^2-8x+5)

Lösung

\int 4 x^{2} - 8 x + 5 d x

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

Schritte zur Lösung

\int 4 x^{2} - 8 x + 5 d x

Vervollständige das Quadrat

4 x^{2} - 8 x + 5 : 4 (x - 1)^{2} + 1

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= \int \frac{1}{2} sec^{3} (v) d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{1}{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) ∣

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int x cos (6 π) x^{2} d x

\int (cos (x) sin (x))^{2} d x

\int \frac{1}{2 1 - 3 v} d v

\int sech^{2} (3 x) d x

\int 4 cos^{3} (x) sin (x) d x