What is the integral of (3x-6)/(4x^2-7x+3) ?

The integral of (3x-6)/(4x^2-7x+3) is

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integral of (3x-6)/(4x^2-7x+3)

\int \frac{3 x - 6}{4 x ^{2} - 7 x + 3} d x

\frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)) + 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣) + C

Solution steps

\int \frac{3 x - 6}{4 x ^{2} - 7 x + 3} d x

Expand

\frac{3 x - 6}{4 x ^{2} - 7 x + 3} : \frac{3 x}{4 x ^{2} - 7 x + 3} - \frac{6}{4 x ^{2} - 7 x + 3}

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{3 x}{4 x ^{2} - 7 x + 3} d x - \int \frac{6}{4 x ^{2} - 7 x + 3} d x

\int \frac{3 x}{4 x ^{2} - 7 x + 3} d x = \frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{ln ∣ 8 x - 6 ∣}{2} - \frac{ln ∣ 8 x - 8 ∣}{2}))

\int \frac{6}{4 x ^{2} - 7 x + 3} d x = - 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)

= \frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{ln ∣ 8 x - 6 ∣}{2} - \frac{ln ∣ 8 x - 8 ∣}{2})) - (- 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣))

Simplify

\frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{ln ∣ 8 x - 6 ∣}{2} - \frac{ln ∣ 8 x - 8 ∣}{2})) - (- 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)) : \frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)) + 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)

= \frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)) + 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)

Add a constant to the solution

= \frac{3}{8} (ln 64 x^{2} - 112 x + 48 - 14 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣)) + 12 (\frac{1}{2} ln ∣ 8 x - 6 ∣ - \frac{1}{2} ln ∣ 8 x - 8 ∣) + C

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What is the integral of (3x-6)/(4x^2-7x+3) ?
The integral of (3x-6)/(4x^2-7x+3) is