What is the integral from 0 to 8 of 1/(x^2-6x+5) ?

The integral from 0 to 8 of 1/(x^2-6x+5) is diverges

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integral from 0 to 8 of 1/(x^2-6x+5)

\int_{0}^{8} \frac{1}{x ^{2} - 6 x + 5} d x

diverges

Solution steps

\int_{0}^{8} \frac{1}{x ^{2} - 6 x + 5} d x

Find undefined (singularity) points:

x = 5, x = 1

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_{0}^{1} \frac{1}{x ^{2} - 6 x + 5} d x + \int_{1}^{5} \frac{1}{x ^{2} - 6 x + 5} d x + \int_{5}^{8} \frac{1}{x ^{2} - 6 x + 5} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (sec (5 x^{3} e^{x}))

x \to 0 lim (6 cos (6 x))

n = 1 \sum \infty \frac{3 ^{n}}{n ! \cdot n}

x \to \infty lim (\frac{1}{x ^{\frac{7}{2}}})

y^{3} - (2 x + 2) + 3 x y^{2} y^{^{'}} = 0

What is the integral from 0 to 8 of 1/(x^2-6x+5) ?
The integral from 0 to 8 of 1/(x^2-6x+5) is diverges