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

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

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

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

ln ∣ x - 1 ∣

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 1}^{x - 1}

Compute the boundaries:

ln ∣ x - 1 ∣

= ln ∣ x - 1 ∣

\int_{- 2}^{3} x - (x^{2} - 6) d x

\int_{0}^{\infty} x e^{- k x} d x

\int_{1}^{3} x^{2} + 8 d x

\int_{x}^{1} x^{2} u^{2} + 1 d u

\int_{1}^{2} 2 x ln (3 x) d x

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