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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{- 2}^{0} \frac{1}{u} d u

= \int_{- 2}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 2}^{0}

Compute the boundaries:

diverges

= diverges

\int_{1}^{3} (2 - x)^{5} d x

\int_{2}^{3} x - 2 d x

\int_{1}^{2} 2^{4 x} d x

\int_{- 1}^{5} 3 x^{2} + 4 x d x

\int_{0}^{2} x^{3} 9 x^{4} + 1 d x

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