What is the integral of (x-1)/(3x^2-6x+5) ?

The integral of (x-1)/(3x^2-6x+5) is 1/6 ln|3x^2-6x+5|+C

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

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

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

Solution steps

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

Expand

\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}

Apply the Sum Rule:

\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))

Add similar elements:

\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

Add a constant to the solution

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

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What is the integral of (x-1)/(3x^2-6x+5) ?
The integral of (x-1)/(3x^2-6x+5) is 1/6 ln|3x^2-6x+5|+C