What is the integral from 0 to 7 of 1/((x-3)^2) ?

The integral from 0 to 7 of 1/((x-3)^2) is diverges

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integral from 0 to 7 of 1/((x-3)^2)

\int_{0}^{7} \frac{1}{( x - 3 ) ^{2}} d x

diverges

Solution steps

\int_{0}^{7} \frac{1}{( x - 3 ) ^{2}} d x

Find undefined (singularity) points:

x = 3

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_{0}^{3} \frac{1}{( x - 3 ) ^{2}} d x + \int_{3}^{7} \frac{1}{( x - 3 ) ^{2}} d x

\int_{0}^{3} \frac{1}{( x - 3 ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{2} x \frac{3}{7} x^{2} d x

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\int_{0}^{2} 9 x^{2} - e^{x} d x

\int_{0}^{1} \frac{x}{1 + 3 x ^{2}} d x

\int_{5}^{\infty} e^{- x} d x

What is the integral from 0 to 7 of 1/((x-3)^2) ?
The integral from 0 to 7 of 1/((x-3)^2) is diverges