integral of (x-1)/(3x^2-6x+5)

Lösung

\int \frac{x - 1}{3 x ^{2} - 6 x + 5} d x

\frac{1}{6} ln 3 x^{2} - 6 x + 5 + C

Schritte zur Lösung

\int \frac{x - 1}{3 x ^{2} - 6 x + 5} d x

Multipliziere aus

\frac{x - 1}{3 x ^{2} - 6 x + 5} : \frac{x}{3 x ^{2} - 6 x + 5} - \frac{1}{3 x ^{2} - 6 x + 5}

Wende die Summenregel an:

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

= \int \frac{x}{3 x ^{2} - 6 x + 5} d x - \int \frac{1}{3 x ^{2} - 6 x + 5} d x

\int \frac{x}{3 x ^{2} - 6 x + 5} d x = \frac{1}{6} ln 3 x^{2} - 6 x + 5 + \frac{1}{6} arctan (\frac{3}{2} (x - 1))

\int \frac{1}{3 x ^{2} - 6 x + 5} d x = \frac{1}{6} arctan (\frac{3}{2} (x - 1))

= \frac{1}{6} ln 3 x^{2} - 6 x + 5 + \frac{1}{6} arctan (\frac{3}{2} (x - 1)) - \frac{1}{6} arctan (\frac{3}{2} (x - 1))

Addiere gleiche Elemente:

\frac{1}{6} arctan (\frac{3}{2} (x - 1)) - \frac{1}{6} arctan (\frac{3}{2} (x - 1)) = 0

= \frac{1}{6} ln 3 x^{2} - 6 x + 5

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln 3 x^{2} - 6 x + 5 + C