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

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

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integral of u/(u-1)

\int \frac{u}{u - 1} d u

u - 1 + ln ∣ u - 1 ∣ + C

Solution steps

\int \frac{u}{u - 1} d u

Apply u-substitution

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

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

Expand

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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= v + ln ∣ v ∣

Substitute back

v = u - 1

= u - 1 + ln ∣ u - 1 ∣

Add a constant to the solution

= u - 1 + ln ∣ u - 1 ∣ + C

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