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

The integral of (x-2)/(x(x+3)(x-4)) is -5/21 ln|x+3|+1/14 ln|x-4|+1/6 ln|x|+C

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

\int \frac{x - 2}{x ( x + 3 ) ( x - 4 )} d x

- \frac{5}{21} ln ∣ x + 3 ∣ + \frac{1}{14} ln ∣ x - 4 ∣ + \frac{1}{6} ln ∣ x ∣ + C

Solution steps

\int \frac{x - 2}{x ( x + 3 ) ( x - 4 )} d x

Expand

\frac{x - 2}{x ( x + 3 ) ( x - 4 )} : \frac{x}{x ( x + 3 ) ( x - 4 )} - \frac{2}{x ( x + 3 ) ( x - 4 )}

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{x}{x ( x + 3 ) ( x - 4 )} d x - \int \frac{2}{x ( x + 3 ) ( x - 4 )} d x

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

\int \frac{2}{x ( x + 3 ) ( x - 4 )} d x = 2 (\frac{1}{21} ln ∣ x + 3 ∣ - \frac{1}{12} ln ∣ x ∣ + \frac{1}{28} ln ∣ x - 4 ∣)

= - \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x - 4 ∣ - 2 (\frac{1}{21} ln ∣ x + 3 ∣ - \frac{1}{12} ln ∣ x ∣ + \frac{1}{28} ln ∣ x - 4 ∣)

Simplify

- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x - 4 ∣ - 2 (\frac{1}{21} ln ∣ x + 3 ∣ - \frac{1}{12} ln ∣ x ∣ + \frac{1}{28} ln ∣ x - 4 ∣) : - \frac{5}{21} ln ∣ x + 3 ∣ + \frac{1}{14} ln ∣ x - 4 ∣ + \frac{1}{6} ln ∣ x ∣

= - \frac{5}{21} ln ∣ x + 3 ∣ + \frac{1}{14} ln ∣ x - 4 ∣ + \frac{1}{6} ln ∣ x ∣

Add a constant to the solution

= - \frac{5}{21} ln ∣ x + 3 ∣ + \frac{1}{14} ln ∣ x - 4 ∣ + \frac{1}{6} ln ∣ x ∣ + C

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What is the integral of (x-2)/(x(x+3)(x-4)) ?
The integral of (x-2)/(x(x+3)(x-4)) is -5/21 ln|x+3|+1/14 ln|x-4|+1/6 ln|x|+C