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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 3

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}^{3} \frac{1}{x ^{2} - 6 x + 9} d x + \int_{3}^{5} \frac{1}{x ^{2} - 6 x + 9} d x

\int_{0}^{3} \frac{1}{x ^{2} - 6 x + 9} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{6} x (6 - x) d x

\int_{1}^{5} k (5 - x) d x

\int_{0}^{2} (x + 1)^{\frac{2}{3}} d x

\int_{2}^{4} - 4 x^{3} ln (x) d x

\int_{0}^{\frac{π}{2}} sin (x^{3}) d x

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