What is the integral from 0 to 3 of 1/(x^2-6x+5) ?

The integral from 0 to 3 of 1/(x^2-6x+5) is diverges

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integral from 0 to 3 of 1/(x^2-6x+5)

\int_{0}^{3} \frac{1}{x ^{2} - 6 x + 5} d x

diverges

Solution steps

\int_{0}^{3} \frac{1}{x ^{2} - 6 x + 5} d x

Find undefined (singularity) points:

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}^{3} \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{6 x}{( x ^{2} + 3 ) ^{2}}

y = 7^{x + a r c c o s (x)}

x^{2} - 4 x, 2 x - x^{2}

\frac{\partial}{\partial y} (x e^{y - z^{2}})

\frac{d}{d x} (ln (2 + e^{x}))

What is the integral from 0 to 3 of 1/(x^2-6x+5) ?
The integral from 0 to 3 of 1/(x^2-6x+5) is diverges