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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

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

\int \frac{3}{x - 3} d x = 3 ln ∣ x - 3 ∣

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

= 3 ln ∣ x - 2 ∣ - \frac{1}{x - 2} - 3 ln ∣ x - 3 ∣ - \frac{2}{x - 3}

Add a constant to the solution

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

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