What is the integral from 3 to 6 of (4x-4)/(x^2-5x) ?

The integral from 3 to 6 of (4x-4)/(x^2-5x) is diverges

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integral from 3 to 6 of (4x-4)/(x^2-5x)

\int_{3}^{6} \frac{4 x - 4}{x ^{2} - 5 x} d x

diverges

Solution steps

\int_{3}^{6} \frac{4 x - 4}{x ^{2} - 5 x} d x

Find undefined (singularity) points:

x = 5

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{3}^{5} \frac{4 x - 4}{x ^{2} - 5 x} d x + \int_{5}^{6} \frac{4 x - 4}{x ^{2} - 5 x} d x

\int_{3}^{5} \frac{4 x - 4}{x ^{2} - 5 x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 3 to 6 of (4x-4)/(x^2-5x) ?
The integral from 3 to 6 of (4x-4)/(x^2-5x) is diverges