What is the integral from 2 to 5 of 1/((x-2)) ?

The integral from 2 to 5 of 1/((x-2)) is diverges

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integral from 2 to 5 of 1/((x-2))

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

diverges

Solution steps

\int_{2}^{5} \frac{1}{x - 2} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{u} d u

= \int_{0}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{3}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{2}} sec (x) - tan (x) d x

\int_{x}^{\infty} \frac{5}{x ^{5}} d x

\int_{1.2}^{1.5} \frac{2}{3} - \frac{2}{9} x d x

\int_{0}^{2 π} 2 - \frac{t}{π} d t

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

What is the integral from 2 to 5 of 1/((x-2)) ?
The integral from 2 to 5 of 1/((x-2)) is diverges