What is the integral of (3x-7)/(x(x^2-7x+12)) ?

The integral of (3x-7)/(x(x^2-7x+12)) is

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integral of (3x-7)/(x(x^2-7x+12))

\int \frac{3 x - 7}{x ( x ^{2} - 7 x + 12 )} d x

- 6 (\frac{1}{2} ln ∣ 2 x - 6 ∣ - \frac{1}{2} ln ∣ 2 x - 8 ∣) - 7 (\frac{1}{12} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + \frac{1}{4} ln ∣ x - 4 ∣) + C

Solution steps

\int \frac{3 x - 7}{x ( x ^{2} - 7 x + 12 )} d x

Expand

\frac{3 x - 7}{x ( x ^{2} - 7 x + 12 )} : \frac{3 x}{x ( x ^{2} - 7 x + 12 )} - \frac{7}{x ( x ^{2} - 7 x + 12 )}

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}{x ( x ^{2} - 7 x + 12 )} d x - \int \frac{7}{x ( x ^{2} - 7 x + 12 )} d x

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

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

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

Add a constant to the solution

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

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What is the integral of (3x-7)/(x(x^2-7x+12)) ?
The integral of (3x-7)/(x(x^2-7x+12)) is