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

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

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

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

diverges

Solution steps

\int_{0}^{5} \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}^{5} \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

x \to 3 + lim (ln x^{2} - 9)

\frac{d ^{3}}{d x ^{3}} (6 x + 8)

(1 - e^{t} + 3 e^{- 4 t}) cos (5 t)

\int \frac{x sin ( x ) + cos ( x ) - 1}{x ^{2}} d x

\frac{\partial}{\partial x} (ln (x^{2} + 3 x y + 2 y^{2}))

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