What is the integral from 1 to 9 of 1/(x-7) ?

The integral from 1 to 9 of 1/(x-7) is diverges

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integral from 1 to 9 of 1/(x-7)

\int_{1}^{9} \frac{1}{x - 7} d x

diverges

Solution steps

\int_{1}^{9} \frac{1}{x - 7} d x

Find undefined (singularity) points:

x = 7

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}^{7} \frac{1}{x - 7} d x + \int_{7}^{9} \frac{1}{x - 7} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (- x e^{(\frac{x + 1}{x - 1})})

\int \frac{x ^{2}}{x ^{2} - 4 x + 5} d x

\frac{d}{d x} (x + 5 e^{x})

\frac{d}{d x} (cot^{3} (4 x))

What is the integral from 1 to 9 of 1/(x-7) ?
The integral from 1 to 9 of 1/(x-7) is diverges