integral of ((2x-1))/(x^2-5x+6)

Lösung

\int \frac{( 2 x - 1 )}{x ^{2} - 5 x + 6} d x

ln 4 x^{2} - 20 x + 24 - 4 ln ∣ 2 x - 4 ∣ + 4 ln ∣ 2 x - 6 ∣ + C

Schritte zur Lösung

\int \frac{2 x - 1}{x ^{2} - 5 x + 6} d x

Vervollständige das Quadrat

x^{2} - 5 x + 6 : (x - \frac{5}{2})^{2} - \frac{1}{4}

= \int \frac{2 x - 1}{( x - \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

Wende die Summenregel an:

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

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

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

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

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

Setze in

u = x - \frac{5}{2}

ein

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

Vereinfache

4 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 1 - ln 2 (x - \frac{5}{2}) + 1 + ln 2 (x - \frac{5}{2}) - 1) : ln 4 x^{2} - 20 x + 24 - 4 ln ∣ 2 x - 4 ∣ + 4 ln ∣ 2 x - 6 ∣

= ln 4 x^{2} - 20 x + 24 - 4 ln ∣ 2 x - 4 ∣ + 4 ln ∣ 2 x - 6 ∣

Füge eine Konstante zur Lösung hinzu

= ln 4 x^{2} - 20 x + 24 - 4 ln ∣ 2 x - 4 ∣ + 4 ln ∣ 2 x - 6 ∣ + C