What is the integral of (2x-6)/(4x^2-24x+32) ?

The integral of (2x-6)/(4x^2-24x+32) is 1/4 ln|4x^2-24x+32|+C

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integral of (2x-6)/(4x^2-24x+32)

\int \frac{2 x - 6}{4 x ^{2} - 24 x + 32} d x

\frac{1}{4} ln 4 x^{2} - 24 x + 32 + C

Solution steps

\int \frac{2 x - 6}{4 x ^{2} - 24 x + 32} d x

Expand

\frac{2 x - 6}{4 x ^{2} - 24 x + 32} : \frac{2 x}{4 x ^{2} - 24 x + 32} - \frac{6}{4 x ^{2} - 24 x + 32}

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{2 x}{4 x ^{2} - 24 x + 32} d x - \int \frac{6}{4 x ^{2} - 24 x + 32} d x

\int \frac{2 x}{4 x ^{2} - 24 x + 32} d x = 2 (\frac{1}{8} ln 4 x^{2} - 24 x + 32 - \frac{3}{8} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣))

\int \frac{6}{4 x ^{2} - 24 x + 32} d x = - \frac{3}{4} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣)

= 2 (\frac{1}{8} ln 4 x^{2} - 24 x + 32 - \frac{3}{8} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣)) - (- \frac{3}{4} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣))

Simplify

2 (\frac{1}{8} ln 4 x^{2} - 24 x + 32 - \frac{3}{8} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣)) - (- \frac{3}{4} (ln ∣ x - 2 ∣ - ln ∣ x - 4 ∣)) : \frac{1}{4} ln 4 x^{2} - 24 x + 32

= \frac{1}{4} ln 4 x^{2} - 24 x + 32

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} - 24 x + 32 + C

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What is the integral of (2x-6)/(4x^2-24x+32) ?
The integral of (2x-6)/(4x^2-24x+32) is 1/4 ln|4x^2-24x+32|+C