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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 2

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x, 0, \frac{9}{x}, x = 5

\frac{d}{d x} (23 arctan (x))

\int \frac{17 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

\int cos^{6} (θ) sin (θ) d θ

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

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