What is the integral of (x-2)/((x-1)(x+2)(x+4)) ?

The integral of (x-2)/((x-1)(x+2)(x+4)) is 2/3 ln|x+2|-3/5 ln|x+4|-1/15 ln|x-1|+C

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integral of (x-2)/((x-1)(x+2)(x+4))

\int \frac{x - 2}{( x - 1 ) ( x + 2 ) ( x + 4 )} d x

\frac{2}{3} ln ∣ x + 2 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ - \frac{1}{15} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x - 2}{( x - 1 ) ( x + 2 ) ( x + 4 )} d x

Expand

\frac{x - 2}{( x - 1 ) ( x + 2 ) ( x + 4 )} : \frac{x}{( x - 1 ) ( x + 2 ) ( x + 4 )} - \frac{2}{( x - 1 ) ( x + 2 ) ( 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 - 1 ) ( x + 2 ) ( x + 4 )} d x - \int \frac{2}{( x - 1 ) ( x + 2 ) ( x + 4 )} d x

\int \frac{x}{( x - 1 ) ( x + 2 ) ( x + 4 )} d x = \frac{1}{15} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{2}{5} ln ∣ x + 4 ∣

\int \frac{2}{( x - 1 ) ( x + 2 ) ( x + 4 )} d x = 2 (\frac{1}{15} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x + 4 ∣)

= \frac{1}{15} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{2}{5} ln ∣ x + 4 ∣ - 2 (\frac{1}{15} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x + 4 ∣)

Simplify

\frac{1}{15} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{2}{5} ln ∣ x + 4 ∣ - 2 (\frac{1}{15} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x + 4 ∣) : \frac{2}{3} ln ∣ x + 2 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ - \frac{1}{15} ln ∣ x - 1 ∣

= \frac{2}{3} ln ∣ x + 2 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ - \frac{1}{15} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ x + 2 ∣ - \frac{3}{5} ln ∣ x + 4 ∣ - \frac{1}{15} ln ∣ x - 1 ∣ + C

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What is the integral of (x-2)/((x-1)(x+2)(x+4)) ?
The integral of (x-2)/((x-1)(x+2)(x+4)) is 2/3 ln|x+2|-3/5 ln|x+4|-1/15 ln|x-1|+C