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

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

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

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

- \frac{1}{x - 2} + \frac{1}{x - 1} + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} ( u - 1 ) ^{2}} d u

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

Expand

\frac{2 u - 1}{u ^{2} ( u - 1 ) ^{2}} : \frac{2 u}{u ^{2} ( u - 1 ) ^{2}} - \frac{1}{u ^{2} ( u - 1 ) ^{2}}

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

\int \frac{2 u}{u ^{2} ( u - 1 ) ^{2}} d u = 2 (ln ∣ u ∣ - ln ∣ u - 1 ∣ - \frac{1}{u - 1})

\int \frac{1}{u ^{2} ( u - 1 ) ^{2}} d u = 2 ln ∣ u ∣ - \frac{1}{u} - 2 ln ∣ u - 1 ∣ - \frac{1}{u - 1}

= 2 (ln ∣ u ∣ - ln ∣ u - 1 ∣ - \frac{1}{u - 1}) - (2 ln ∣ u ∣ - \frac{1}{u} - 2 ln ∣ u - 1 ∣ - \frac{1}{u - 1})

Substitute back

u = x - 1

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

Simplify

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

= - \frac{1}{x - 2} + \frac{1}{x - 1}

Add a constant to the solution

= - \frac{1}{x - 2} + \frac{1}{x - 1} + C

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