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

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

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

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

ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - \frac{1}{u}

Substitute back

u = x - 1

= ln ∣ x - 1 ∣ - \frac{1}{x - 1}

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

\int (\frac{1}{x} - \frac{5}{9 x ^{2} + 1}) d x

\frac{d}{d x} (3 x^{2} y)

x \to 2 lim (x^{3} + 2 x^{2} - x - 4)

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

2 x^{2} e^{x}

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