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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{u} d u

= \int_{0}^{1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{\partial}{\partial w} (\frac{w}{h ( w ) ( w ) ^{2}})

\frac{d}{d x} (e^{- x} (4 - x) - 2)

\frac{6 s + 2}{s ( 9 s ^{2} + 6 s + 2 )}

\frac{d}{d x} (\frac{2 x}{125})

\frac{6}{x ^{2} + 2}

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