What is the integral of (x-8)/(x(x^2-4x+4)) ?

The integral of (x-8)/(x(x^2-4x+4)) is 3/(x-2)-2ln|x|+2ln|x-2|+C

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integral of (x-8)/(x(x^2-4x+4))

\int \frac{x - 8}{x ( x ^{2} - 4 x + 4 )} d x

\frac{3}{x - 2} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 8}{x ( x ^{2} - 4 x + 4 )} d x

Expand

\frac{x - 8}{x ( x ^{2} - 4 x + 4 )} : \frac{x}{x ( x ^{2} - 4 x + 4 )} - \frac{8}{x ( x ^{2} - 4 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 ^{2} - 4 x + 4 )} d x - \int \frac{8}{x ( x ^{2} - 4 x + 4 )} d x

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

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

= - \frac{1}{x - 2} - 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )})

Simplify

- \frac{1}{x - 2} - 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )}) : \frac{3}{x - 2} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣

= \frac{3}{x - 2} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{3}{x - 2} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣ + C

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What is the integral of (x-8)/(x(x^2-4x+4)) ?
The integral of (x-8)/(x(x^2-4x+4)) is 3/(x-2)-2ln|x|+2ln|x-2|+C