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

The integral of ((x^2-2x))/((x-1)^2) is x+1/(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 + \frac{1}{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 + \frac{1}{x - 1} - 1

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

Add a constant to the solution

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

\int x^{6} ln (x) d x

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\frac{\partial}{\partial z} (x e^{\frac{y}{z}})

\frac{d}{d x} (e^{c o s (x)} \cdot sin (x^{2} - x))

2 x^{2}, 2 x + 6, - 1.3, 2.3

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