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

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

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

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

x + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 1 + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

= x + 4 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 1 + C

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