What is the integral from 1 to 12 of 1/(x-9) ?

The integral from 1 to 12 of 1/(x-9) is diverges

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integral from 1 to 12 of 1/(x-9)

\int_{1}^{12} \frac{1}{x - 9} d x

diverges

Solution steps

\int_{1}^{12} \frac{1}{x - 9} d x

Find undefined (singularity) points:

x = 9

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}^{9} \frac{1}{x - 9} d x + \int_{9}^{12} \frac{1}{x - 9} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{1}{n + 4}

\frac{\partial}{\partial x} (x \cdot e^{x} + ln (x) y)

\int - 7 ln (x^{2} - 1) d x

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

\frac{d y}{d x}, 4 y^{2} - 5 y + 8 = 2 x^{2} - 3 x

What is the integral from 1 to 12 of 1/(x-9) ?
The integral from 1 to 12 of 1/(x-9) is diverges