What is the integral from 6 to 13 of 1/((x-8)^2) ?

The integral from 6 to 13 of 1/((x-8)^2) is diverges

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integral from 6 to 13 of 1/((x-8)^2)

\int_{6}^{13} \frac{1}{( x - 8 ) ^{2}} d x

diverges

Solution steps

\int_{6}^{13} \frac{1}{( x - 8 ) ^{2}} d x

Find undefined (singularity) points:

x = 8

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_{6}^{8} \frac{1}{( x - 8 ) ^{2}} d x + \int_{8}^{13} \frac{1}{( x - 8 ) ^{2}} d x

\int_{6}^{8} \frac{1}{( x - 8 ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x y^{^{'}} - y = \frac{3}{4} x ln (x)

\frac{\partial}{\partial y} (2 x y)

\frac{d y}{d x}, x y = 24

\frac{d}{d r} (4 π r^{2})

What is the integral from 6 to 13 of 1/((x-8)^2) ?
The integral from 6 to 13 of 1/((x-8)^2) is diverges