integral of sqrt(3-2x+x^2)

Lösung

\int 3 - 2 x + x^{2} d x

\frac{1}{2} (x - 1) x^{2} - 2 x + 3 + ln \frac{1}{2} (x - 1) + \frac{1}{2} (x^{2} - 2 x + 3) + C

Schritte zur Lösung

\int 3 - 2 x + x^{2} d x

Vervollständige das Quadrat

3 - 2 x + x^{2} : (x - 1)^{2} + 2

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Vereinfache

2 \frac{sec ^{2} ( arctan ( \frac{1}{2} ( x - 1 ) ) ) sin ( arctan ( \frac{1}{2} ( x - 1 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x - 1))) + sec (arctan (\frac{1}{2} (x - 1))) : \frac{1}{2} (x - 1) x^{2} - 2 x + 3 + ln \frac{1}{2} (x - 1) + \frac{1}{2} (x^{2} - 2 x + 3)

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x - 1) x^{2} - 2 x + 3 + ln \frac{1}{2} (x - 1) + \frac{1}{2} (x^{2} - 2 x + 3) + C

in v erse l a pl a ce ⎩ ⎨ ⎧ (\frac{( s + 1 ) ^{3}}{4})^{2} ⎭ ⎬ ⎫

\frac{d}{d x} (c x^{3})

t an g e n t f (x) = x^{4} + 2 e^{x}, at x = 0

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

\int_{0}^{4} (x) d x