What is the integral from 3 to 5 of 1/(x-3) ?

The integral from 3 to 5 of 1/(x-3) is diverges

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integral from 3 to 5 of 1/(x-3)

\int_{3}^{5} \frac{1}{x - 3} d x

diverges

Solution steps

\int_{3}^{5} \frac{1}{x - 3} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\int_{0}^{2} x^{2} + 4 d x

\int_{0}^{\frac{π}{6}} cos^{6} (3 x) d x

\int_{1}^{e} (ln (x))^{9} d x

\int_{0}^{1} (1 - x^{2})^{5} d x

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

What is the integral from 3 to 5 of 1/(x-3) ?
The integral from 3 to 5 of 1/(x-3) is diverges