What is the integral from 1 to 7 of 1/(x-2) ?

The integral from 1 to 7 of 1/(x-2) is diverges

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integral from 1 to 7 of 1/(x-2)

\int_{1}^{7} \frac{1}{x - 2} d x

diverges

Solution steps

\int_{1}^{7} \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}^{7} \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

\int \frac{x ^{3}}{x ^{2} + 9} d x

4^{- 3} (\frac{2}{s ^{7}} + 3 s)^{4}

n = 0 \sum \infty \frac{1}{2} (\frac{1}{2})^{n}

\int \frac{2 x + 3}{x ^{2} + 1} d x

x \to 4 - lim (16 - x^{2})

What is the integral from 1 to 7 of 1/(x-2) ?
The integral from 1 to 7 of 1/(x-2) is diverges