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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{4} \frac{1}{u} d u

= \int_{0}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{4}

Compute the boundaries:

diverges

= diverges

\int_{- 1}^{0} x cos (nπ x) d x

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

\int_{- 1}^{1} x^{4} sin (x) d x

\int_{0}^{\infty} x cos (x) d x

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

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