What is the integral of (3x^2-6x+4)/((x-1)(x-2)^2) ?

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

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integral of (3x^2-6x+4)/((x-1)(x-2)^2)

\int \frac{3 x ^{2} - 6 x + 4}{( x - 1 ) ( x - 2 ) ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2} - 6 x + 4}{( x - 1 ) ( x - 2 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} - 6 x + 4}{( x - 1 ) ( x - 2 ) ^{2}} : \frac{1}{x - 1} + \frac{2}{x - 2} + \frac{4}{( x - 2 ) ^{2}}

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

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

\int \frac{2}{x - 2} d x = 2 ln ∣ x - 2 ∣

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

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

Add a constant to the solution

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

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