integral of ((3x-5))/(sqrt(2x^2-6x+3))

Lösung

\int \frac{( 3 x - 5 )}{2 x ^{2} - 6 x + 3} d x

2 (\frac{3}{4} 4 x^{2} - 12 x + 6 - \frac{1}{4} ln \frac{4 x ^{2} - 12 x + 6}{3} + \frac{2}{3} x - 3) + C

Schritte zur Lösung

\int \frac{3 x - 5}{2 x ^{2} - 6 x + 3} d x

Vervollständige das Quadrat

2 x^{2} - 6 x + 3 : 2 (x - \frac{3}{2})^{2} - \frac{3}{2}

= \int \frac{3 x - 5}{2 ( x - \frac{3}{2} ) ^{2} - \frac{3}{2}} d x

Wende U-Substitution an

= \int \frac{2 ( 3 ( u + \frac{3}{2} ) - 5 )}{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{3 ( u + \frac{3}{2} ) - 5}{4 u ^{2} - 3} d u

Multipliziere aus

\frac{3 ( u + \frac{3}{2} ) - 5}{4 u ^{2} - 3} : \frac{3 u}{4 u ^{2} - 3} - \frac{1}{2 4 u ^{2} - 3}

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

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

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

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

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

Setze in

u = x - \frac{3}{2}

ein

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

Vereinfache

2 \frac{3}{4} 4 (x - \frac{3}{2})^{2} - 3 - \frac{1}{4} ln \frac{1}{3} (4 (x - \frac{3}{2})^{2} - 3) + \frac{2}{3} (x - \frac{3}{2}) : 2 (\frac{3}{4} 4 x^{2} - 12 x + 6 - \frac{1}{4} ln \frac{4 x ^{2} - 12 x + 6}{3} + \frac{2}{3} x - 3)

= 2 (\frac{3}{4} 4 x^{2} - 12 x + 6 - \frac{1}{4} ln \frac{4 x ^{2} - 12 x + 6}{3} + \frac{2}{3} x - 3)

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{3}{4} 4 x^{2} - 12 x + 6 - \frac{1}{4} ln \frac{4 x ^{2} - 12 x + 6}{3} + \frac{2}{3} x - 3) + C