What is the integral from 0 to 1/2 of 1/(x-1) ?

The integral from 0 to 1/2 of 1/(x-1) is -ln(2)

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integral from 0 to 1/2 of 1/(x-1)

\int_{0}^{\frac{1}{2}} \frac{1}{x - 1} d x

- ln (2)

Decimal Notation

- 0.69314 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{x - 1} d x

Apply u-substitution

: \int_{- 1}^{- \frac{1}{2}} \frac{1}{u} d u

= \int_{- 1}^{- \frac{1}{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 1}^{- \frac{1}{2}}

Compute the boundaries:

- ln (2)

= - ln (2)

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\int_{- 0.5}^{0} x ln (x + 1) d x

What is the integral from 0 to 1/2 of 1/(x-1) ?
The integral from 0 to 1/2 of 1/(x-1) is -ln(2)