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

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

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integral of-1/(1-x)

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

ln ∣ 1 - x ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

= - (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- ln ∣ u ∣)

Substitute back

u = 1 - x

= - (- ln ∣ 1 - x ∣)

Simplify

= ln ∣ 1 - x ∣

Add a constant to the solution

= ln ∣ 1 - x ∣ + C

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