What is the integral from 7 to 14 of 1/((x-9)^2) ?

The integral from 7 to 14 of 1/((x-9)^2) is diverges

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integral from 7 to 14 of 1/((x-9)^2)

\int_{7}^{14} \frac{1}{( x - 9 ) ^{2}} d x

diverges

Solution steps

\int_{7}^{14} \frac{1}{( x - 9 ) ^{2}} 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_{7}^{9} \frac{1}{( x - 9 ) ^{2}} d x + \int_{9}^{14} \frac{1}{( x - 9 ) ^{2}} d x

\int_{7}^{9} \frac{1}{( x - 9 ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{4} ∣ 3 t - 8 ∣ d t

f (x) = \frac{4 x ^{2} - x + 3}{x}

\int \frac{e ^{- s t}}{- s} d t

\frac{s ^{4} + 10}{s - 2}

\frac{\partial}{\partial x} (6 x^{2} + 5 x^{3} y^{4} - 10 y^{5})

What is the integral from 7 to 14 of 1/((x-9)^2) ?
The integral from 7 to 14 of 1/((x-9)^2) is diverges