What is the integral of (3w-2)/(3w^2-4w+1) ?

The integral of (3w-2)/(3w^2-4w+1) is 1/2 ln|3w^2-4w+1|+C

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integral of (3w-2)/(3w^2-4w+1)

\int \frac{3 w - 2}{3 w ^{2} - 4 w + 1} d w

\frac{1}{2} ln 3 w^{2} - 4 w + 1 + C

Solution steps

\int \frac{3 w - 2}{3 w ^{2} - 4 w + 1} d w

Apply u-substitution

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 3 w^{2} - 4 w + 1

= \frac{1}{2} ln 3 w^{2} - 4 w + 1

Add a constant to the solution

= \frac{1}{2} ln 3 w^{2} - 4 w + 1 + C

What is the integral of (3w-2)/(3w^2-4w+1) ?
The integral of (3w-2)/(3w^2-4w+1) is 1/2 ln|3w^2-4w+1|+C