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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

2 (x - 1 + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{3}{x - 1} : 2 x - \frac{1}{x - 1} - 2

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

Add a constant to the solution

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

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