What is the integral from 4 to 6 of 1/(x-4) ?

The integral from 4 to 6 of 1/(x-4) is diverges

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integral from 4 to 6 of 1/(x-4)

\int_{4}^{6} \frac{1}{x - 4} d x

diverges

Solution steps

\int_{4}^{6} \frac{1}{x - 4} 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

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

\int_{0}^{1} x^{2} e^{- 13 x} d x

\int_{0}^{1} \frac{x - 4}{x ^{2} + 5 x + 6} d x

\int_{0}^{4} x (x - 1) d x

What is the integral from 4 to 6 of 1/(x-4) ?
The integral from 4 to 6 of 1/(x-4) is diverges