What is the integral of 1/((x-3)(x^2-2x)^2) ?

The integral of 1/((x-3)(x^2-2x)^2) is 1/9 ln|x-3|-1/9 ln|x|+1/(12x)+1/(4(x-2))+C

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integral of 1/((x-3)(x^2-2x)^2)

\int \frac{1}{( x - 3 ) ( x ^{2} - 2 x ) ^{2}} d x

\frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )} + C

Solution steps

\int \frac{1}{( x - 3 ) ( x ^{2} - 2 x ) ^{2}} d x

Take the partial fraction of

\frac{1}{( x - 3 ) ( x ^{2} - 2 x ) ^{2}} : \frac{1}{9 ( x - 3 )} - \frac{1}{9 x} - \frac{1}{12 x ^{2}} - \frac{1}{4 ( x - 2 ) ^{2}}

= \int \frac{1}{9 ( x - 3 )} - \frac{1}{9 x} - \frac{1}{12 x ^{2}} - \frac{1}{4 ( x - 2 ) ^{2}} d x

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{1}{9 ( x - 3 )} d x - \int \frac{1}{9 x} d x - \int \frac{1}{12 x ^{2}} d x - \int \frac{1}{4 ( x - 2 ) ^{2}} d x

\int \frac{1}{9 ( x - 3 )} d x = \frac{1}{9} ln ∣ x - 3 ∣

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{12 x ^{2}} d x = - \frac{1}{12 x}

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

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ - (- \frac{1}{12 x}) - (- \frac{1}{4 ( x - 2 )})

Simplify

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )}

Add a constant to the solution

= \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{12 x} + \frac{1}{4 ( x - 2 )} + C

What is the integral of 1/((x-3)(x^2-2x)^2) ?
The integral of 1/((x-3)(x^2-2x)^2) is 1/9 ln|x-3|-1/9 ln|x|+1/(12x)+1/(4(x-2))+C