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

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

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integral from 0 to 5 of 1/(x-3)

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

diverges

Solution steps

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

\int_{0}^{3} \frac{1}{x - 3} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

θ \to 0 lim (\frac{θ}{sin ^{2} ( θ )})

x \to 2 lim (x^{2} - 7 x + 12)

\frac{1}{( s ^{3} + s ^{2} + s )}

x \to 0 lim (\frac{9 sin ( 1 )}{1})

\frac{d y}{d x} + 2 x y = 9 x

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