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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 4}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\infty} e^{- m x} d x

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

\int_{0}^{2} \frac{x}{50} d x

\int_{- 4}^{4} x^{3} d x

\int_{- 1}^{0} x^{2} e^{x^{3}} d x

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