What is the integral from 0 to 9 of 1/(x^2-6x+5) ?

The integral from 0 to 9 of 1/(x^2-6x+5) is diverges

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integral from 0 to 9 of 1/(x^2-6x+5)

\int_{0}^{9} \frac{1}{x ^{2} - 6 x + 5} d x

diverges

Solution steps

\int_{0}^{9} \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}^{9} \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{1}{1 + x ^{2}}, \frac{x ^{2}}{2}

\frac{\partial}{\partial x} (y arctan (\frac{x}{z}))

e^{2 x^{2}}

x \to \infty lim (\frac{7.5 x + 120000}{x})

\int \frac{4}{4 + e ^{4 x}} d x

What is the integral from 0 to 9 of 1/(x^2-6x+5) ?
The integral from 0 to 9 of 1/(x^2-6x+5) is diverges