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

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

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

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

- 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) - 2 b (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) + C

Solution steps

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

Expand

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

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

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

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

= - 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) - 2 b (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣))

Add a constant to the solution

= - 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) - 2 b (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) + C

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