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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} - 6 x + 2

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

Add a constant to the solution

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

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