What is the integral from a to b of 1/(1-x) ?

The integral from a to b of 1/(1-x) is -ln|1-b|+ln|1-a|

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integral from a to b of 1/(1-x)

\int_{a}^{b} \frac{1}{1 - x} d x

- ln ∣ 1 - b ∣ + ln ∣ 1 - a ∣

Solution steps

\int_{a}^{b} \frac{1}{1 - x} d x

Apply u-substitution

: \int_{1 - a}^{1 - b} - \frac{1}{u} d u

= \int_{1 - a}^{1 - b} - \frac{1}{u} d u

Take the constant out:

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

= - \int_{1 - a}^{1 - b} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{1 - a}^{1 - b}

Compute the boundaries:

ln ∣ 1 - b ∣ - ln ∣ 1 - a ∣

= - (ln ∣ 1 - b ∣ - ln ∣ 1 - a ∣)

Simplify

= - ln ∣ 1 - b ∣ + ln ∣ 1 - a ∣

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What is the integral from a to b of 1/(1-x) ?
The integral from a to b of 1/(1-x) is -ln|1-b|+ln|1-a|