What is the integral of (x^2-5x+3)/((x-1)(x-2)^2) ?

The integral of (x^2-5x+3)/((x-1)(x-2)^2) is 2ln|x-2|+3/(x-2)-ln|x-1|+C

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integral of (x^2-5x+3)/((x-1)(x-2)^2)

\int \frac{x ^{2} - 5 x + 3}{( x - 1 ) ( x - 2 ) ^{2}} d x

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

Solution steps

\int \frac{x ^{2} - 5 x + 3}{( x - 1 ) ( x - 2 ) ^{2}} d x

Apply u-substitution

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

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

Expand

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

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

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

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

= ln ∣ u - 1 ∣ - \frac{1}{u - 1} - (- \frac{3}{u - 1}) - (ln ∣ u ∣ - ln ∣ u - 1 ∣ - \frac{1}{u - 1})

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x^2-5x+3)/((x-1)(x-2)^2) ?
The integral of (x^2-5x+3)/((x-1)(x-2)^2) is 2ln|x-2|+3/(x-2)-ln|x-1|+C