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

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

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

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

ln ∣ u - 1 ∣ - \frac{1}{u - 1} + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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}{v} d v + \int \frac{1}{v ^{2}} d v

\int \frac{1}{v} d v = ln ∣ v ∣

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

= ln ∣ v ∣ - \frac{1}{v}

Substitute back

v = u - 1

= ln ∣ u - 1 ∣ - \frac{1}{u - 1}

Add a constant to the solution

= ln ∣ u - 1 ∣ - \frac{1}{u - 1} + C

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