integral of sqrt(x^2-x+1)

Lösung

\int x^{2} - x + 1 d x

\frac{1}{8} (2 (2 x - 1) x^{2} - x + 1 + 3 (ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 - \frac{ln ( 3 )}{2})) + C

Schritte zur Lösung

\int x^{2} - x + 1 d x

Vervollständige das Quadrat

x^{2} - x + 1 : (x - \frac{1}{2})^{2} + \frac{3}{4}

= \int (x - \frac{1}{2})^{2} + \frac{3}{4} d x

Wende U-Substitution an

= \int \frac{4 u ^{2} + 3}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int 4 u^{2} + 3 d u

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

Ersetze zurück

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

Vereinfache

\frac{1}{2} \cdot \frac{3}{2} \frac{sec ^{2} ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2}))) : \frac{1}{8} (2 (2 x - 1) x^{2} - x + 1 + 3 (ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 - \frac{ln ( 3 )}{2}))

= \frac{1}{8} (2 (2 x - 1) x^{2} - x + 1 + 3 (ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 - \frac{ln ( 3 )}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (2 (2 x - 1) x^{2} - x + 1 + 3 (ln 2 x + 3 \frac{4 x ^{2} - 4 x + 4}{3} - 1 - \frac{ln ( 3 )}{2})) + C

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