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

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

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integral of 1/(1-u)

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

- ln ∣ 1 - u ∣ + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

v = 1 - u

= - ln ∣ 1 - u ∣

Add a constant to the solution

= - ln ∣ 1 - u ∣ + C