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

Lösung

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

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

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

\frac{2 u - 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} - \frac{1}{4 u ^{2} + 3}

Wende die Summenregel an:

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

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

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

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2 3} arctan (\frac{2}{3} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} u))

Setze in

u = x - \frac{1}{2}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

x \to - \infty lim (e^{x} - 1)

x \to \frac{π}{4} lim (\frac{sec ^{2} ( x ) - 2 tan ( x )}{1 + cos ( 4 x )})

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{9}{x ^{2}}) y = 0

x \to - 2 + lim (\frac{- 2}{x ^{2} - 4})

x \to + 0 lim (\frac{1}{x})