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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \int 1 + \frac{2}{u} + \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u + \int \frac{2}{u} d u + \int \frac{1}{u ^{2}} d u

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= u + 2 ln ∣ u ∣ - \frac{1}{u}

Substitute back

u = x - 1

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

Add a constant to the solution

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

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

\int \frac{1}{x ^{a}} d x

x \to - 1 + lim (\frac{x}{x + 1})

\frac{\partial}{\partial t} (3 t)

\frac{d y}{d x} = \frac{x + y - 2}{2 x + 2 y + 3}

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