What is the integral from 0 to 5 of 1/(x-5) ?

The integral from 0 to 5 of 1/(x-5) is diverges

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integral from 0 to 5 of 1/(x-5)

\int_{0}^{5} \frac{1}{x - 5} d x

diverges

Solution steps

\int_{0}^{5} \frac{1}{x - 5} d x

Apply u-substitution

: \int_{- 5}^{0} \frac{1}{u} d u

= \int_{- 5}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 5}^{0}

Compute the boundaries:

diverges

= diverges

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

\int_{1}^{7} \frac{1}{8 x + 8} d x

\int_{0}^{1} 5 - (- \frac{3}{2} x + 4) d x

\int_{0}^{\frac{8}{3}} 3 t - 8 d t

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

What is the integral from 0 to 5 of 1/(x-5) ?
The integral from 0 to 5 of 1/(x-5) is diverges