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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (\frac{x ^{2} + 9}{x})

\frac{d}{d x} (x^{2} ln^{2} (x))

n = 0 \sum \infty e^{- \frac{5 n}{2}}

\frac{\partial}{\partial x} (x^{3} e^{y} + cos (y))

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

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