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

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

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integral from 2 to 5 of 1/(x-3)

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

diverges

Solution steps

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

\int_{2}^{3} \frac{1}{x - 3} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\int ln (x + x^{2}) d x

\int x^{2} (7 - x^{3})^{\frac{2}{3}} d x

\frac{d}{d x} (tan (x^{2} - 4 x))

x \to - \infty lim (x^{3} - 5 x^{2})

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