What is the integral of (6x-3)/(4x^2-4x+1) ?

The integral of (6x-3)/(4x^2-4x+1) is 3/4 ln|4x^2-4x+1|+C

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integral of (6x-3)/(4x^2-4x+1)

\int \frac{6 x - 3}{4 x ^{2} - 4 x + 1} d x

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

Solution steps

\int \frac{6 x - 3}{4 x ^{2} - 4 x + 1} d x

Apply u-substitution

: \int \frac{3}{4 u} d u

= \int \frac{3}{4 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{3}{4} ln ∣ u ∣

Substitute back

u = 4 x^{2} - 4 x + 1

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

Add a constant to the solution

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

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What is the integral of (6x-3)/(4x^2-4x+1) ?
The integral of (6x-3)/(4x^2-4x+1) is 3/4 ln|4x^2-4x+1|+C